Thin Lenses, Optical Instruments, and Defects of Vision Alıştırma Soruları — JAMB UTME

Match each defect of vision to its correct image formation characteristic and mode of optical correction.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Hypermetropia pairs with light focusing behind the retina (corrected with a converging lens). Myopia pairs with light focusing in front of the retina (corrected with a diverging lens). Presbyopia pairs with age-related loss of accommodation (corrected with bifocal/converging lenses). Astigmatism pairs with unequal corneal curvature in different planes (corrected with a cylindrical lens).

Each vision defect uniquely corresponds to a specific optical fault and corrective lens: Hypermetropia focuses images behind the retina (convex lens), Myopia focuses images in front of the retina (concave lens), Presbyopia involves age-related elasticity loss (bifocal/convex lens), and Astigmatism stems from asymmetric corneal curvature (cylindrical lens).

Adım Adım Çözüm

1

Identify the structural defect and focus point for Hypermetropia.

Near point moves farther away, so light rays from standard near point focus behind the retina. A converging (convex) lens is required to bend light rays inward prior to entering the eye.

Hypermetropic eyes lack sufficient refractive power for near vision.

2

Identify the structural defect and focus point for Myopia.

Far point is reduced, so light rays from distant objects focus in front of the retina. A diverging (concave) lens is required to diverge incoming parallel rays.

Myopic eyes possess excessive refractive power or an elongated eyeball.

3

Identify the cause and correction for Presbyopia.

Caused by aging of ciliary muscles and hardening of the eye lens. Corrected with converging or bifocal lenses.

Presbyopia specifically relates to loss of elastic accommodation in older individuals.

4

Identify the cause and correction for Astigmatism.

Caused by non-uniform curvature of the cornea along horizontal vs vertical axes. Corrected using cylindrical lenses.

Cylindrical lenses compensate for asymmetrical focal lengths along specific axes.

Anahtar Kavram

Defects of Vision and Corrective Lenses

Soru 2Soru →

An object is placed $12\text{ cm}$ in front of a diverging lens having a focal length of $18\text{ cm}$ . What is the magnitude of the image distance from the lens?

7.2\text{ cm}

36.0\text{ cm}

30.0\text{ cm}

6.0\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $7.2\text{ cm}$

Cevap

The magnitude of the image distance is

7.2\text{ cm}

.

By the thin lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

, substituting

f = -18\text{ cm}

(due to the lens being diverging) and

u = +12\text{ cm}

gives

\frac{1}{v} = -\frac{1}{18} - \frac{1}{12} = -\frac{5}{36}\text{ cm}^{-1}

, which yields

v = -7.2\text{ cm}

. Thus, the image is located

7.2\text{ cm}

from the lens.

Adım Adım Çözüm

1

Identify the given values and assign the correct signs according to the real-is-positive sign convention.

Object distance

u = +12\text{ cm}

; Focal length of diverging (concave) lens

f = -18\text{ cm}

.

Diverging lenses have virtual focus, so their focal length must be taken as negative.

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

to solve for image distance

v

.

\frac{1}{-18} = \frac{1}{12} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{18} - \frac{1}{12}

.

Rearranging the equation isolates the reciprocal of the image distance.

3

Find a common denominator and solve for

v

.

\frac{1}{v} = \frac{-2 - 3}{36} = -\frac{5}{36}\text{ cm}^{-1} \implies v = -\frac{36}{5} = -7.2\text{ cm}

.

The negative sign indicates that the image formed is virtual and located on the same side of the lens as the object.

4

Take the magnitude of the image distance.

|v| = 7.2\text{ cm}

.

The question asks for the distance, which is a scalar magnitude.

Anahtar Kavram

Thin Lens Formula & Sign Convention for Concave Lenses

Tahmini Süre:1m 30s

Soru 3Soru →

An object is placed $10\text{ cm}$ in front of a converging lens of focal length $15\text{ cm}$ . What is the distance of the image from the lens and its nature?

30\text{ cm}

from the lens and virtual

30\text{ cm}

from the lens and real

6\text{ cm}

from the lens and real

6\text{ cm}

from the lens and virtual

Cevabı ve açıklamayı göster

Cevap: $30\text{ cm}$ from the lens and virtual

Cevap

The image is

30\text{ cm}

from the lens and virtual.

Using the thin lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with

f = +15\text{ cm}

and

u = +10\text{ cm}

yields

\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30}\text{ cm}^{-1}

. Taking the reciprocal gives

v = -30\text{ cm}

. The magnitude of the image distance is

30\text{ cm}

, and the negative sign confirms that the image formed is virtual.

Adım Adım Çözüm

1

Identify the given optical parameters and sign conventions.

Object distance

u = +10\text{ cm}

, Focal length for converging lens

f = +15\text{ cm}

.

By convention for real objects and converging lenses, both

u

and

f

are positive.

2

Substitute the values into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

.

\frac{1}{15} = \frac{1}{10} + \frac{1}{v}

The thin lens formula relates focal length, object distance, and image distance.

3

Rearrange to solve for the image distance reciprocal

\frac{1}{v}

.

\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}\text{ cm}^{-1}

Subtracting

\frac{1}{10}

from

\frac{1}{15}

requires finding a common denominator of 30.

4

Take the reciprocal to find

v

and interpret its sign.

v = -30\text{ cm}

The magnitude of the distance is

30\text{ cm}

. The negative sign indicates that the image is formed on the same side as the object, making it virtual.

Anahtar Kavram

Thin lens formula and sign convention for image formation

Tahmini Süre:1m 30s

Soru 4Soru →

Match each eye defect to its corresponding optical cause and method of vision correction.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Myopia corresponds to image formation in front of the retina (corrected with a concave lens); Hypermetropia corresponds to image formation behind the retina (corrected with a convex lens); Astigmatism corresponds to uneven corneal curvature (corrected with a cylindrical lens); Presbyopia corresponds to age-related loss of accommodation (corrected with bifocal lenses).

Each vision defect is matched precisely to its biological cause and optical correction method: Myopia with image in front of retina and concave lens, Hypermetropia with image behind retina and convex lens, Astigmatism with non-spherical cornea and cylindrical lens, and Presbyopia with age-related loss of accommodation and bifocal lenses.

Adım Adım Çözüm

1

Identify the cause and corrective lens for Myopia.

Myopia causes light to focus in front of the retina due to excessive converging power or an elongated eyeball, requiring a concave (diverging) lens.

A concave lens diverges incoming light rays prior to entry into the eye so the focal point moves back to the retina.

2

Identify the cause and corrective lens for Hypermetropia.

Hypermetropia causes light to focus behind the retina due to insufficient converging power or a shortened eyeball, requiring a convex (converging) lens.

A convex lens provides extra converging power to bring the focal point forward onto the retina.

3

Identify the cause and corrective lens for Astigmatism.

Astigmatism arises from irregular corneal curvature, requiring a cylindrical lens.

Cylindrical lenses vary focal power along a specific axis to equalize uneven light refraction.

4

Identify the cause and corrective lens for Presbyopia.

Presbyopia is caused by loss of elasticity in the lens with age, requiring bifocal lenses.

Bifocal lenses provide dual correction zones for distance and near vision.

Anahtar Kavram

Vision Defects and Lens Corrections

Soru 5Soru →

A compound microscope in normal adjustment consists of an objective lens with a focal length of $1.5\text{ cm}$ and an eyepiece with a focal length of $5.0\text{ cm}$ . An object is placed at a distance of $1.6\text{ cm}$ in front of the objective lens. Calculate the distance, in centimeters, between the objective lens and the eyepiece.

Cevabı ve açıklamayı göster

Cevap: 29

Cevap

The distance between the objective lens and the eyepiece is 29.0 cm.

Applying the thin lens formula to the objective lens yields

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o}

, giving an image distance

v_o = 24.0\text{ cm}

. Under normal adjustment, the intermediate image falls on the focal point of the eyepiece, making

u_e = f_e = 5.0\text{ cm}

. The total separation between the two lenses is

L = v_o + u_e = 24.0 + 5.0 = 29.0\text{ cm}

.

Adım Adım Çözüm

1

Apply the thin lens formula to the objective lens to find the intermediate image position

v_o

.

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o} \Rightarrow \frac{1}{v_o} = \frac{2}{3} - \frac{5}{8} = \frac{1}{24}\text{ cm}^{-1} \Rightarrow v_o = 24.0\text{ cm}

The objective lens forms a real, inverted, magnified image at distance

v_o

from the objective.

2

Identify the object distance for the eyepiece

u_e

under normal adjustment.

u_e = f_e = 5.0\text{ cm}

For normal adjustment of a optical instrument, the final image is formed at infinity, requiring the intermediate image to sit exactly at the principal focus of the eyepiece.

3

Sum the intermediate image distance and eyepiece object distance to obtain total lens separation

L

.

L = v_o + u_e = 24.0\text{ cm} + 5.0\text{ cm} = 29.0\text{ cm}

The separation of lenses in a compound microscope is the distance from the objective to the intermediate image plus the distance from the intermediate image to the eyepiece.

Anahtar Kavram

Compound microscope optics and lens separation in normal adjustment

Soru 6Soru →

A converging lens has a focal length of $25\text{ cm}$ . What is the power of the lens in dioptres?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The power of the lens is

4\text{ D}

.

The power

P

of a lens in dioptres is defined as the reciprocal of its focal length

f

expressed in meters (

P = \frac{1}{f}

). Expressing

25\text{ cm}

in meters gives

0.25\text{ m}

. Substituting this value yields

P = \frac{1}{0.25\text{ m}} = 4\text{ D}

.

Adım Adım Çözüm

1

Convert the focal length from centimeters to meters

f = 25\text{ cm} = 0.25\text{ m}

The unit of dioptre (

D

) is defined as reciprocal meters (

m^{-1}

), so focal length must be in meters.

2

Apply the lens power formula

P = \frac{1}{f}

and compute the value

P = \frac{1}{0.25} = 4\text{ D}

The power of a converging lens is positive and equal to the reciprocal of its focal length in meters.

Anahtar Kavram

Power of a Lens

Soru 7Soru →

A hypermetropic eye has its unassisted near point situated at $75\text{ cm}$ from the eye. Calculate the focal length, in $\text{cm}$ , of the converging spectacle lens needed to enable the person to read a book comfortably at a distance of $25\text{ cm}$ from the eye.

Cevabı ve açıklamayı göster

Cevap: 37.5

Cevap

The focal length of the required converging lens is

37.5\text{ cm}

.

To correct hypermetropia, the spectacle lens must form a virtual image of an object located at the desired near point (

u = +25\text{ cm}

) at the eye's actual, unassisted near point (

v = -75\text{ cm}

). Substituting

u = +25\text{ cm}

and

v = -75\text{ cm}

into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

gives

\frac{1}{f} = \frac{1}{25} - \frac{1}{75} = \frac{2}{75}\text{ cm}^{-1}

. Taking the reciprocal yields

f = 37.5\text{ cm}

.

Adım Adım Çözüm

1

Determine the object distance and image distance with sign conventions

u = +25\text{ cm}

and

v = -75\text{ cm}

.

The object is placed at the normal reading distance (

25\text{ cm}

), and the spectacle lens creates a virtual image on the same side of the lens at the person's near point (

75\text{ cm}

).

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{25} - \frac{1}{75}.

The thin lens formula relates the focal length to object and image distances.

3

Calculate the focal length

f

f = 37.5\text{ cm}.

Simplifying

\frac{3 - 1}{75} = \frac{2}{75}

and taking the reciprocal yields

f = \frac{75}{2} = 37.5\text{ cm}

.

Anahtar Kavram

Correction of hypermetropia (farsightedness) using thin lens formula with virtual image sign convention

Soru 8Soru →

An astronomical telescope in normal adjustment consists of an objective lens with a focal length of $60\text{ cm}$ and an eyepiece with a focal length of $5\text{ cm}$ . What is the angular magnification produced by the telescope and the separation distance between the two lenses?

12

and

65\text{ cm}

0.083

and

65\text{ cm}

12

and

55\text{ cm}

300

and

65\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $12$ and $65\text{ cm}$

Cevap

The angular magnification is

12

and the separation distance between the lenses is

65\text{ cm}

.

For an astronomical telescope in normal adjustment, the magnification is given by

M = \frac{f_o}{f_e} = \frac{60}{5} = 12

, and the length of the telescope tube (distance between lenses) is

L = f_o + f_e = 60 + 5 = 65\text{ cm}

.

Adım Adım Çözüm

1

Calculate the angular magnification (

M

) of the telescope

M = \frac{f_o}{f_e} = \frac{60}{5} = 12

In an astronomical telescope in normal adjustment, angular magnification is given by the ratio of the focal length of the objective lens to that of the eyepiece.

2

Calculate the separation distance (

L

) between the two lenses

L = f_o + f_e = 60 + 5 = 65\text{ cm}

When in normal adjustment, the final image is formed at infinity, so the distance between the objective lens and eyepiece equals the sum of their focal lengths.

Anahtar Kavram

Astronomical telescope in normal adjustment (magnification and lens separation)

Tahmini Süre:1m 0s

Soru 9Soru →

A converging lens of focal length $20\text{ cm}$ is placed in thin coaxial contact with a diverging lens of focal length $50\text{ cm}$ . An object is placed $30\text{ cm}$ in front of this lens combination. What is the position and nature of the final image formed?

300\text{ cm}

in front of the combination (virtual image)

60\text{ cm}

behind the combination (real image)

27.3\text{ cm}

behind the combination (real image)

15.8\text{ cm}

behind the combination (real image)

Cevabı ve açıklamayı göster

Cevap: $300\text{ cm}$ in front of the combination (virtual image)

Cevap

The image is virtual and formed

300\text{ cm}

in front of the lens combination.

Combining a converging lens (

f = +20\text{ cm}

) and a diverging lens (

f = -50\text{ cm}

) yields an effective focal length of

F = +\frac{100}{3}\text{ cm}

. Using the lens formula

\frac{1}{F} = \frac{1}{u} + \frac{1}{v}

with object distance

u = 30\text{ cm}

gives

\frac{1}{v} = \frac{3}{100} - \frac{1}{30} = -\frac{1}{300}\text{ cm}^{-1}

, resulting in

v = -300\text{ cm}

. The negative sign confirms the image is virtual and located

300\text{ cm}

in front of the combination.

Adım Adım Çözüm

1

Calculate the effective focal length (

F

) of the two lenses in contact.

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{20\text{ cm}} + \frac{1}{-50\text{ cm}} = \frac{5 - 2}{100\text{ cm}} = \frac{3}{100}\text{ cm}^{-1}

, so

F = +\frac{100}{3}\text{ cm}

.

Thin lenses in contact combine algebraically according to their optical powers, taking signs into account (positive for converging, negative for diverging).

2

Apply the lens formula to find the image distance (

v

).

\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{30} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{1}{30} = \frac{9 - 10}{300} = -\frac{1}{300}\text{ cm}^{-1}

.

Rearranging the thin lens equation allows us to solve for the image distance

v

given object distance

u = 30\text{ cm}

.

3

Interpret the sign and magnitude of

v

.

v = -300\text{ cm}

, which signifies a virtual image located

300\text{ cm}

in front of the lens combination (on the object side).

A negative image distance in the standard real-is-positive convention denotes a virtual image.

Anahtar Kavram

Combination of thin lenses in contact and lens sign conventions

Tahmini Süre:2m 0s

Soru 10Soru →

A myopic person has a far point of $52\text{ cm}$ from the eye. A corrective lens is placed $2\text{ cm}$ in front of the eye to enable the person to see distant objects clearly. A second thin converging lens of focal length $+20\text{ cm}$ is then placed in thin coaxial contact with this corrective lens. If an object is placed $50\text{ cm}$ in front of this combined lens system, what is the position and nature of the final image formed?

100\text{ cm}

behind the combined lens (real)

20\text{ cm}

behind the combined lens (real)

100\text{ cm}

in front of the combined lens (virtual)

20\text{ cm}

in front of the combined lens (virtual)

Cevabı ve açıklamayı göster

Cevap: $100\text{ cm}$ behind the combined lens (real)

Cevap

100\text{ cm}

behind the combined lens (real image)

The corrective lens for short-sightedness (myopia) must be a diverging lens with a negative focal length. Accounting for the

2\text{ cm}

distance between the eye and the lens, the far point relative to the lens is

50\text{ cm}

, giving

f_1 = -50\text{ cm}

. Combining this lens with the converging lens (

f_2 = +20\text{ cm}

) yields a net power

\frac{1}{F} = -\frac{1}{50} + \frac{1}{20} = +\frac{3}{100}\text{ cm}^{-1}

, or

F = +\frac{100}{3}\text{ cm}

. Placing an object at

u = 50\text{ cm}

gives

\frac{1}{v} = \frac{3}{100} - \frac{1}{50} = +\frac{1}{100}\text{ cm}^{-1}

, which yields

v = +100\text{ cm}

. The positive sign confirms a real image formed

100\text{ cm}

behind the lens system.

Adım Adım Çözüm

1

Calculate the focal length

f_1

of the corrective lens required for myopia.

f_1 = -50\text{ cm}

The far point distance from the lens is

52\text{ cm} - 2\text{ cm} = 50\text{ cm}

. A diverging (concave) lens is needed to form a virtual image of distant objects (

u = \infty

) at the far point (

v = -50\text{ cm}

).

2

Determine the focal length

F

of the two lenses in thin contact.

F = +\frac{100}{3}\text{ cm}

Using the combined focal length equation

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = -\frac{1}{50} + \frac{1}{20} = \frac{-2 + 5}{100} = +\frac{3}{100}\text{ cm}^{-1}

.

3

Apply the thin lens formula to locate the final image distance

v

for

u = 50\text{ cm}

.

v = +100\text{ cm}

Using

\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{50} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{2}{100} = \frac{1}{100}\text{ cm}^{-1}

.

4

Determine the nature of the image from the sign of

v

.

Real image formed

100\text{ cm}

behind the combined lens system.

A positive value of image distance (

v > 0

) indicates a real image formed on the opposite side (behind) the lens system.

Anahtar Kavram

Thin lens combination and sight defect correction sign conventions

Tahmini Süre:3m 0s

Soru 11Soru →

A short-sighted person cannot see objects clearly beyond a distance of $80\text{ cm}$ . What type of lens, of what focal length and power, is required to correct this vision defect so that the person can view distant objects clearly?

Diverging lens of focal length

80\text{ cm}

and power

-1.25\text{ D}

Converging lens of focal length

80\text{ cm}

and power

+1.25\text{ D}

Converging lens of focal length

36.4\text{ cm}

and power

+2.75\text{ D}

Diverging lens of focal length

80\text{ cm}

and power

-0.0125\text{ D}

Cevabı ve açıklamayı göster

Cevap: Diverging lens of focal length $80\text{ cm}$ and power $-1.25\text{ D}$

Cevap

Diverging lens of focal length

80\text{ cm}

and power

-1.25\text{ D}

For a myopic person with a far point at

80\text{ cm}

, parallel rays from a distant object (

u = \infty

) must be diverged so they appear to come from the far point (

v = -80\text{ cm} = -0.8\text{ m}

). Using

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

, we obtain

f = -0.8\text{ m} = -80\text{ cm}

. The negative focal length corresponds to a diverging lens, and its power is

P = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}

.

Adım Adım Çözüm

1

Identify the optical requirements for correcting myopia (short-sightedness).

For a distant object (

u = \infty

), the corrective lens must form a virtual image at the eye's far point (

v = -80\text{ cm} = -0.8\text{ m}

).

Myopic eyes focus rays from infinity in front of the retina; placing a virtual image at the far point allows the eye lens to focus it correctly onto the retina.

2

Apply the thin lens formula to calculate the required focal length.

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{\infty} + \frac{1}{-0.8\text{ m}} = 0 - 1.25\text{ m}^{-1} \implies f = -0.8\text{ m} = -80\text{ cm}$.

The negative sign indicates that a concave (diverging) lens is required.

3

Calculate the power of the corrective lens in dioptres.

$P = \frac{1}{f\text{ (in metres)}} = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}$.

Lens power in dioptres (D) is the reciprocal of the focal length expressed in meters.

Anahtar Kavram

Correction of Myopia (Short-Sightedness) using Diverging Lenses

Soru 12Soru →

A thin converging lens forms a real image of an object on a screen placed $60\text{ cm}$ from the lens. If the object is located $30\text{ cm}$ in front of the lens, what is the focal length of the lens in centimeters ( $\text{cm}$ )?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The focal length of the converging lens is

20\text{ cm}

.

Using the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with an object distance

u = 30\text{ cm}

and a real image distance

v = 60\text{ cm}

yields

\frac{1}{f} = \frac{1}{30} + \frac{1}{60} = \frac{1}{20}

, giving

f = 20\text{ cm}

.

Adım Adım Çözüm

1

Identify given parameters and apply correct sign conventions

u = +30\text{ cm}

(real object) and

v = +60\text{ cm}

(real image on screen)

In thin lens calculations for real objects and images formed on screens, both distances are positive.

2

Substitute parameters into the thin lens formula

\frac{1}{f} = \frac{1}{30} + \frac{1}{60}

The thin lens equation relates focal length

f

, object distance

u

, and image distance

v

.

3

Perform fraction addition and solve for focal length

\frac{1}{f} = \frac{3}{60} = \frac{1}{20} \implies f = 20\text{ cm}

Taking the common denominator gives

\frac{1}{20}\text{ cm}^{-1}

, which yields

f = 20\text{ cm}

.

Anahtar Kavram

Thin Lens Formula for Real Image Formation

Soru 13Soru →

Match each visual defect or optical condition listed in Column A with its corresponding cause and corrective lens in Column B.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Myopia pairs with rays focusing in front of the retina (diverging lens correction); Hypermetropia pairs with rays focusing behind the retina (converging lens correction); Astigmatism pairs with uneven corneal curvature (cylindrical lens correction); Presbyopia pairs with age-related loss of accommodation (bifocal lens correction).

Each defect of vision is accurately matched to its optical cause and standard corrective device: Myopia uses diverging lenses to push the image focal plane onto the retina, Hypermetropia uses converging lenses to pull the image forward onto the retina, Astigmatism uses cylindrical lenses for non-spherical corneal curves, and Presbyopia uses bifocal/converging lenses to correct age-related accommodation loss.

Adım Adım Çözüm

1

Analyze Myopia

Myopia causes distant rays to focus before reaching the retina because the eye lens is overly converging or the eye focal length is too short; a diverging (concave) lens spreads rays to push the focal point back onto the retina.

Identify optical cause and lens remedy for short-sightedness.

2

Analyze Hypermetropia

Hypermetropia causes near rays to focus behind the retina; a converging (convex) lens bends incoming light rays inwards to bring the focal point onto the retina.

Identify optical cause and lens remedy for long-sightedness.

3

Analyze Astigmatism

Astigmatism arises from non-uniform curvature of the refracting surfaces, requiring a cylindrical lens with differential curvature along different planes.

Identify refractive error causing multiple focal planes.

4

Analyze Presbyopia

Presbyopia is due to age-induced stiffening of the eye lens and loss of ciliary accommodation power, which is managed using bifocal or converging lenses.

Distinguish physiological aging effects on focal accommodation.

Anahtar Kavram

Defects of Vision and Corrective Lenses

Tahmini Süre:1m 30s

Soru 14Soru →

Match each defect of vision on the left with its corresponding corrective optical lens on the right.

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Cevabı ve açıklamayı göster

Cevap

Myopia matches with Concave (diverging) lens, Hypermetropia matches with Convex (converging) lens, Presbyopia matches with Bifocal lens, and Astigmatism matches with Cylindrical lens.

Each eye condition is paired with its specific optical correction: Myopia requires a concave lens to diverge light, Hypermetropia requires a convex lens to converge light, Presbyopia uses a bifocal lens to assist diminished accommodation, and Astigmatism relies on a cylindrical lens to correct asymmetrical curvature.

Adım Adım Çözüm

1

Analyze Myopia (Short-sightedness)

Parallel rays focus in front of the retina due to an elongated eyeball or over-refractive lens.

Diverging (concave) lenses spread out incoming rays slightly before entering the eye so the focal point shifts back onto the retina.

2

Analyze Hypermetropia (Long-sightedness)

Light rays focus behind the retina due to a shortened eyeball or insufficient focal power.

Converging (convex) lenses provide additional converging power to focus rays directly on the retina.

3

Analyze Presbyopia

The eye lens loses elasticity with age, reducing its power to accommodate both near and far objects.

Bifocal lenses have two distinct focal lengths in a single glass unit to assist with both near and distant vision.

4

Analyze Astigmatism

Cornea or crystalline lens curvature is uneven along different axes, producing distorted vision.

Cylindrical lenses correct uneven refractive power by bending light along one axis without affecting the orthogonal axis.

Anahtar Kavram

Defects of Vision and Corrective Lenses

Tahmini Süre:45s

Soru 15Soru →

An object is placed $15.0\text{ cm}$ in front of a thin diverging lens with a focal length of $10.0\text{ cm}$ . What is the image distance formed by the lens?

-6.0\text{ cm}

(

6.0\text{ cm}

on the same side as the object)

+30.0\text{ cm}

(

30.0\text{ cm}

on the opposite side of the object)

+6.0\text{ cm}

(

6.0\text{ cm}

on the opposite side of the object)

-30.0\text{ cm}

(

30.0\text{ cm}

on the same side as the object)

Cevabı ve açıklamayı göster

Cevap: $-6.0\text{ cm}$ ( $6.0\text{ cm}$ on the same side as the object)

Cevap

The image distance is

-6.0\text{ cm}

, indicating a virtual image located

6.0\text{ cm}

in front of the lens on the same side as the object.

Using the thin lens equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with

u = +15.0\text{ cm}

and

f = -10.0\text{ cm}

for the diverging lens gives

\frac{1}{v} = -\frac{1}{10} - \frac{1}{15} = -\frac{1}{6}

, resulting in

v = -6.0\text{ cm}

. The negative sign confirms the image is virtual and formed on the same side as the object.

Adım Adım Çözüm

1

Identify the given optical parameters and apply the proper sign convention.

Object distance

u = +15.0\text{ cm}

; Focal length for a diverging lens

f = -10.0\text{ cm}

.

By the real-is-positive sign convention, real object distance

u

is positive, while the focal length

f

of a concave/diverging lens is strictly negative.

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

and solve for

\frac{1}{v}

.

\frac{1}{-10.0} = \frac{1}{15.0} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{10.0} - \frac{1}{15.0}

Isolating the reciprocal image distance term requires subtracting

\frac{1}{u}

from both sides.

3

Calculate the common denominator and evaluate

v

.

\frac{1}{v} = \frac{-3 - 2}{30.0} = -\frac{5.0}{30.0} = -\frac{1}{6.0} \implies v = -6.0\text{ cm}

Inverting the reciprocal yields the final signed image distance.

Anahtar Kavram

Thin Lens Formula and Sign Conventions for Diverging Lenses

Tahmini Süre:1m 15s

Soru 16Soru →

A hypermetropic (far-sighted) person has a near point located at a distance of $100\text{ cm}$ from the eye. What power of spectacle lens, in dioptres, is required to enable this person to read print comfortably held at the standard near point of $25\text{ cm}$ ?

+3.0\text{ D}

-3.0\text{ D}

+5.0\text{ D}

+0.75\text{ D}

Cevabı ve açıklamayı göster

Cevap: $+3.0\text{ D}$

Cevap

+3.0\text{ D}

To correct hypermetropia, a converging (convex) lens is required to bend incoming rays so that an object placed at the standard near point of

25\text{ cm}

(

+0.25\text{ m}

) forms a virtual image at the defective eye's near point of

100\text{ cm}

(

-1.0\text{ m}

). Substituting

u = +0.25\text{ m}

and

v = -1.0\text{ m}

into the power formula

P = \frac{1}{u} + \frac{1}{v}

yields

P = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

.

Adım Adım Çözüm

1

Identify object distance (

u

) and required virtual image distance (

v

)

u = +25\text{ cm} = +0.25\text{ m}

and

v = -100\text{ cm} = -1.0\text{ m}

The lens must create a virtual image (on the same side as the object) at the person's actual near point when an object is placed at the standard reading distance.

2

Apply the thin lens formula to determine lens power

P

P = \frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Power in dioptres is the reciprocal of the focal length in metres.

3

Calculate the numerical value of lens power

P = \frac{1}{0.25\text{ m}} + \frac{1}{-1.0\text{ m}} = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

Adding the reciprocal quantities yields a positive focal power of

+3.0\text{ D}

.

Anahtar Kavram

Correction of Hypermetropia using Converging Lenses

Tahmini Süre:1m 30s

Soru 17Soru →

Match each optical instrument listed on the left with its corresponding lens configuration and image characteristics on the right.

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Cevabı ve açıklamayı göster

Cevap

Astronomical Telescope pairs with the configuration having

f_o > f_e

forming an image at infinity; Simple Microscope pairs with a single converging lens forming an erect virtual image; Compound Microscope pairs with two converging lenses having

f_o < f_e

; and Projection Lantern pairs with a converging lens forming a real, inverted image on a screen.

Each instrument matches its distinct optical construction: telescopes use

f_o > f_e

for distant viewing at infinity, simple microscopes use a single convex lens for virtual magnifying, compound microscopes use

f_o < f_e

for double magnification of tiny objects, and projectors use a single convex lens to cast real images onto a distant surface.

Adım Adım Çözüm

1

Determine the lens setup and final image position of an astronomical telescope in normal adjustment.

The objective has a larger focal length than the eyepiece (

f_o > f_e

), and the final image is formed at infinity.

Telescopes gather light from distant objects, requiring a larger objective focal length for high angular magnification and comfortable viewing at infinity.

2

Determine the configuration of a simple microscope.

It consists of a single convex lens producing an erect, virtual, and magnified image.

When an object is placed within the focal length of a single convex lens, it acts as a magnifying glass.

3

Determine the focal length relationship of a compound microscope.

It uses two convex lenses where the objective focal length is shorter than the eyepiece focal length (

f_o < f_e

).

A very short objective focal length maximizes linear magnification of small, near objects before the eyepiece further magnifies the intermediate image.

4

Determine the type of image produced by a projection lantern (slide projector).

It forms a real, inverted, and magnified image on a screen.

Projecting images onto a screen requires a real image formed by a converging lens.

Anahtar Kavram

Optical Instrument Lens Configurations and Image Properties

Soru 18Soru →

A converging lens has a focal length of $25\text{ cm}$ . What is the optical power of the lens in dioptres?

+4.0\text{ D}

-4.0\text{ D}

+0.04\text{ D}

+25.0\text{ D}

Cevabı ve açıklamayı göster

Cevap: $+4.0\text{ D}$

Cevap

The optical power of the lens is

+4.0\text{ D}

.

The optical power

P

of a lens in dioptres (

\text{D}

) is calculated using

P = \frac{1}{f}

, where

f

is the focal length in metres. Converting

25\text{ cm}

to metres gives

0.25\text{ m}

. Since the lens is converging, its focal length is positive (

+0.25\text{ m}

). Thus,

P = \frac{1}{+0.25} = +4.0\text{ D}

.

Adım Adım Çözüm

1

Convert the focal length from centimetres to metres

f = 25\text{ cm} = 0.25\text{ m}

The unit of optical power (dioptre,

\text{D}

) requires the focal length to be expressed in metres.

2

Apply the sign convention for a converging lens

f = +0.25\text{ m}

A converging (convex) lens has a real principal focus, so its focal length is positive by sign convention.

3

Calculate the optical power using the formula

P = \frac{1}{f}

P = \frac{1}{+0.25} = +4.0\text{ D}

Optical power is defined as the inverse of the focal length in metres.

Anahtar Kavram

Power of a Thin Lens

Tahmini Süre:45s

Soru 19Soru →

A thin converging lens of focal length $15\text{ cm}$ forms an erect image that is magnified three times. What is the distance of the object from the lens?

10\text{ cm}

20\text{ cm}

5\text{ cm}

45\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $10\text{ cm}$

Cevap

10\text{ cm}

An erect image produced by a thin converging lens is virtual, which means the linear magnification is positive (

m = +3

) and the image distance is negative relative to the real object (

v = -3u

). Substituting

f = +15\text{ cm}

and

v = -3u

into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

gives

\frac{1}{15} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

. Solving for

u

yields

u = 10\text{ cm}

.

Adım Adım Çözüm

1

Determine the nature of the image and establish the relationship between image distance and object distance

Since the image formed by a converging lens is erect, it must be virtual. Thus, linear magnification

m = +3 = -\frac{v}{u}

, giving

v = -3u

.

A single convex lens produces an erect image only when the image is virtual, requiring a negative image distance under standard optical sign conventions.

2

Substitute focal length and image distance into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{15} = \frac{1}{u} + \frac{1}{-3u}

The thin lens equation relates focal length, object distance, and image distance.

3

Simplify the algebraic expression and solve for object distance u

\frac{1}{15} = \frac{3 - 1}{3u} = \frac{2}{3u} \implies 3u = 30 \implies u = 10\text{ cm}

Finding a common denominator allows direct solution for the object distance

u

.

Anahtar Kavram

Thin lens formula and sign conventions for virtual images

Soru 20Soru →

Match each optical instrument component or device on the left with its correct focal length requirement and image formation condition on the right.

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Cevabı ve açıklamayı göster

Cevap

Compound Microscope Objective Lens pairs with very short focal length producing a real, inverted, and magnified intermediate image. Astronomical Telescope Objective Lens pairs with long focal length forming a real, inverted, and diminished image of a distant body. Simple Magnifying Glass pairs with single convex lens with object placed within focal length producing an erect, virtual, and magnified image. Slide / Film Projector Lens pairs with convex lens with object positioned between f and 2f producing a real, inverted, and enlarged image on a screen.

The matching correct pairs align each optical device with its precise optical parameter: microscope objectives utilize short focal lengths for high magnification of near objects, telescope objectives utilize long focal lengths for distant objects, simple magnifiers place objects closer than the focal point to form virtual erect images, and projectors place objects between one and two focal lengths to cast real enlarged images on a screen.

Adım Adım Çözüm

1

Analyze the objective lens of a compound microscope

It requires a very short focal length to achieve high linear magnification of near objects.

The specimen is placed just outside the focal point (

f < u < 2f

), producing a real, inverted, magnified image inside the tube.

2

Analyze the objective lens of an astronomical telescope

It requires a long focal length and large aperture.

Distant celestial bodies subtend tiny angles at the eye; a long focal length objective creates a larger real intermediate image for the eyepiece to magnify.

3

Analyze the ray path condition of a simple magnifying glass

The object is held within the principal focus (

u < f

).

Light rays emerging from the lens diverge, forming an enlarged, virtual, and erect image on the same side of the lens as the object.

4

Analyze the image projection setup in a slide projector

The slide is placed between

f

and

2f

of a converging lens.

According to the thin lens formula, placing an object between

f

and

2f

yields a real, inverted, magnified image beyond

2f

.

Anahtar Kavram

Operating principles and ray placement parameters of optical instruments

Thin Lenses, Optical Instruments, and Defects of Vision

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