Reflection of Light at Plane and Curved Mirrors Alıştırma Soruları — JAMB UTME

A concave mirror has a radius of curvature of $40\text{ cm}$ . What is the focal length of the mirror?

20\text{ cm}

-20\text{ cm}

40\text{ cm}

80\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $20\text{ cm}$

Cevap

The focal length of the concave mirror is

20\text{ cm}

.

For any spherical mirror, the focal length is half of its radius of curvature (

f = \frac{r}{2}

). Since the mirror is concave, its focal length is positive, yielding

\frac{40\text{ cm}}{2} = 20\text{ cm}

.

Adım Adım Çözüm

1

Identify the given physical quantity

Radius of curvature

r = 40\text{ cm}

for a concave mirror.

The radius of curvature is the distance from the pole to the center of curvature.

2

Apply the relationship between focal length and radius of curvature

f = \frac{r}{2} = \frac{40\text{ cm}}{2} = 20\text{ cm}

.

For spherical mirrors, the principal focus lies halfway between the pole and the center of curvature.

Anahtar Kavram

Focal length of spherical mirrors

Soru 2Soru →

A convex security mirror installed in a store has a focal length of magnitude $20\text{ cm}$ . If an upright image of a customer is formed with a linear magnification of $0.25$ , at what distance from the mirror is the customer standing?

60\text{ cm}

100\text{ cm}

15\text{ cm}

25\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $60\text{ cm}$

Cevap

The customer is standing at a distance of

60\text{ cm}

from the convex mirror.

By sign convention, a convex mirror has a negative focal length (

f = -20\text{ cm}

). An upright image formed by a mirror has a positive magnification

m = +0.25

. Using

m = -v/u

, we get

v = -u/4

. Substituting these into the mirror equation

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

gives

-\frac{1}{20} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

, which yields

u = 60\text{ cm}

.

Adım Adım Çözüm

1

Identify the optical properties and apply sign conventions for a convex mirror

Focal length

f = -20\text{ cm}

(convex mirror focal length is virtual/negative). Magnification

m = +0.25

(upright image).

Convex mirrors always form virtual, upright, diminished images behind the mirror.

2

Relate image distance

v

to object distance

u

using the linear magnification formula

Since

m = -v/u = 1/4

, we obtain

v = -u/4

.

The negative sign in the magnification definition accounts for virtual image distance.

3

Substitute

f = -20\text{ cm}

and

v = -u/4

into the mirror formula

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

\frac{1}{-20} = \frac{1}{u} + \frac{1}{-u/4} \implies -\frac{1}{20} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

.

Combines fractions with a common denominator

u

to solve for the unknown object distance.

4

Solve for the object distance

u

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

.

Cross-multiplying yields the real object distance in front of the mirror.

Anahtar Kavram

Reflection at Convex Mirrors and Optical Sign Conventions

Soru 3Soru →

An object is placed $12\text{ cm}$ in front of a concave mirror having a focal length of $20\text{ cm}$ . What is the distance of the image from the mirror and its nature?

30\text{ cm}

behind the mirror, virtual

30\text{ cm}

in front of the mirror, real

7.5\text{ cm}

in front of the mirror, real

7.5\text{ cm}

behind the mirror, virtual

Cevabı ve açıklamayı göster

Cevap: $30\text{ cm}$ behind the mirror, virtual

Cevap

The image is formed

30\text{ cm}

behind the mirror and is virtual.

Applying the mirror formula

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

with

f = 20\text{ cm}

and

u = 12\text{ cm}

yields

\frac{1}{v} = \frac{1}{20} - \frac{1}{12} = -\frac{1}{30}\text{ cm}^{-1}

, resulting in

v = -30\text{ cm}

. The negative sign confirms that the image is virtual and located

30\text{ cm}

behind the mirror.

Adım Adım Çözüm

1

Identify given quantities and apply standard mirror sign conventions.

Object distance

u = +12\text{ cm}

, focal length

f = +20\text{ cm}

.

For a concave mirror, real objects and real focal points carry positive values.

2

Set up the mirror equation

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

.

\frac{1}{20} = \frac{1}{12} + \frac{1}{v}

The mirror formula relates focal length, object distance, and image distance.

3

Rearrange the equation to isolate

\frac{1}{v}

and solve.

\frac{1}{v} = \frac{1}{20} - \frac{1}{12} = \frac{3 - 5}{60} = -\frac{2}{60} = -\frac{1}{30}\text{ cm}^{-1} \implies v = -30\text{ cm}.

Finding a common denominator of 60 allows exact calculation of the negative reciprocal.

4

Determine image characteristics based on the sign of

v

.

The image distance magnitude is

30\text{ cm}

behind the mirror, and the image is virtual.

A negative value for image distance

v

signifies a virtual image located behind the mirror surface.

Anahtar Kavram

Concave Mirror Formula and Virtual Image Formation

Soru 4Soru →

For any real object positioned in front of a convex spherical mirror, the image formed is always virtual, upright, and diminished, located between the pole and the principal focus.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is true. A convex mirror always produces a virtual, erect, and diminished image positioned between the pole and the focus for any real object placed in front of it.

The statement accurately expresses the fundamental optics rule for convex mirrors: reflecting surface geometry causes incident parallel rays to diverge, producing a virtual, erect, and diminished image located between the mirror's pole and principal focus for all real object positions.

Adım Adım Çözüm

1

Analyze ray tracing principles for a convex mirror.

Ray 1, traveling parallel to the principal axis, reflects such that it appears to originate from the principal focus

F

behind the mirror. Ray 2, directed toward the center of curvature

C

, reflects back along its original path.

Tracing these rays determines where their virtual extensions intersect.

2

Determine the location and characteristics of the virtual intersection.

The backward extensions of the reflected rays intersect behind the reflecting surface between the pole

P

and the principal focus

F

.

Since the light rays diverge and only their extensions meet, the image formed is virtual, erect, and diminished.

Anahtar Kavram

Image characteristics of convex spherical mirrors

Soru 5Soru →

A dentist uses a small concave mirror with a focal length of $20\text{ mm}$ to inspect a patient's tooth. If the mirror produces an upright image that is magnified $4$ times, how far from the tooth is the mirror placed?

15\text{ mm}

25\text{ mm}

80\text{ mm}

5\text{ mm}

Cevabı ve açıklamayı göster

Cevap: $15\text{ mm}$

Cevap

The mirror must be placed

15\text{ mm}

from the tooth.

For a concave mirror, an upright image is virtual. Linear magnification

m = 4

implies

v = -4u

. Substituting

f = 20\text{ mm}

into the mirror formula

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

gives

\frac{1}{20} = \frac{1}{u} - \frac{1}{4u} = \frac{3}{4u}

, solving to

u = 15\text{ mm}

.

Adım Adım Çözüm

1

Identify given values and apply sign conventions.

Focal length of concave mirror

f = +20\text{ mm}

. Magnification

m = +4

because the image is upright (virtual).

Upright images formed by single optical mirrors are always virtual, requiring a negative image distance.

2

Express image distance

v

in terms of object distance

u

.

Linear magnification

m = -\frac{v}{u} \implies +4 = -\frac{v}{u} \implies v = -4u

.

The linear magnification formula relates orientation, object distance, and image distance.

3

Substitute

f

and

v

into the mirror formula.

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{20} = \frac{1}{u} - \frac{1}{4u} = \frac{3}{4u}.

The mirror equation determines object and image locations relative to the focal length.

4

Solve for the object distance

u

.

4u = 60 \implies u = 15\text{ mm}.

Cross-multiplying gives the required distance between the mirror and the tooth.

Anahtar Kavram

Reflection and image formation by concave spherical mirrors (virtual magnified image)

Soru 6Soru →

If a light ray undergoes successive reflections from two plane mirrors inclined at an angle $\theta$ to each other in a single plane, the total angle of deviation produced in the ray is independent of the initial angle of incidence at the first mirror.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is TRUE. The total angle of deviation experienced by a ray after two successive reflections at inclined plane mirrors is

\delta = 360^\circ - 2\theta

, which depends only on the angle of inclination

\theta

and is independent of the initial angle of incidence.

The net angular deviation for a ray undergoing two successive reflections at plane mirrors inclined at angle

\theta

is

\delta = 360^\circ - 2\theta

. Since the initial angle of incidence

i_1

cancels out during geometric summation, the total deviation is completely independent of the angle of incidence.

Adım Adım Çözüm

1

Determine the deviation at the first mirror

\delta_1 = 180^\circ - 2i_1

For a single reflection at a plane mirror with angle of incidence

i_1

, the angle of deviation is

\delta_1 = 180^\circ - 2i_1

.

2

Express the angle of incidence at the second mirror in terms of inclination angle

\theta

i_2 = \theta - i_1

From the geometric construction of the ray path inside the triangle formed by the two mirror surfaces, the interior angle relationship gives

i_1 + i_2 = \theta

.

3

Calculate the total deviation after both reflections

\delta = \delta_1 + \delta_2 = (180^\circ - 2i_1) + (180^\circ - 2i_2) = 360^\circ - 2(i_1 + i_2) = 360^\circ - 2\theta

Summing the individual deviations eliminates the variable

i_1

, demonstrating that the total deviation depends only on the inclination angle

\theta

.

Anahtar Kavram

Total Angle of Deviation for Inclined Plane Mirrors

Soru 7Soru →

A convex mirror forms an upright image that is $\frac{1}{3}$ the size of an object. When the object is moved $20\text{ cm}$ further away from the mirror, the size of the image becomes $\frac{1}{5}$ the size of the object. What is the radius of curvature of the mirror?

20\text{ cm}

10\text{ cm}

40\text{ cm}

5\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $20\text{ cm}$

Cevap

The radius of curvature of the convex mirror is

20\text{ cm}

.

For a convex mirror, the focal length is negative (

f = -f_0

). The magnification formula

m = \frac{f}{f - u}

for a virtual upright image gives

m = \frac{-f_0}{-f_0 - u} = \frac{f_0}{f_0 + u}

. For

m_1 = \frac{1}{3}

, we get

u_1 = 2f_0

. For

m_2 = \frac{1}{5}

, we get

u_2 = 4f_0

. The object displacement is

u_2 - u_1 = 2f_0 = 20\text{ cm}

, which gives

f_0 = 10\text{ cm}

. The radius of curvature is

R = 2f_0 = 20\text{ cm}

, making the choice stating

20\text{ cm}

correct.

Adım Adım Çözüm

1

Apply the magnification formula and sign convention for a convex mirror at the first position.

For a convex mirror,

f = -f_0

and the image is virtual (

v_1 = -v_{01}

). Magnification

m_1 = -\frac{v_1}{u_1} = \frac{v_{01}}{u_1} = \frac{1}{3}

, so

v_{01} = \frac{u_1}{3}

.

Convex mirrors always form virtual, upright, and diminished images.

2

Substitute

v_1

into the mirror equation for the first position to express

u_1

in terms of focal length magnitude

f_0

.

\frac{1}{-f_0} = \frac{1}{u_1} + \frac{1}{-v_{01}} = \frac{1}{u_1} - \frac{3}{u_1} = -\frac{2}{u_1} \implies u_1 = 2f_0

.

The mirror formula is

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

with cartesian sign conventions.

3

Repeat the mirror equation calculation for the second object position.

For

m_2 = \frac{1}{5}

,

v_{02} = \frac{u_2}{5}

. Substituting gives

\frac{1}{-f_0} = \frac{1}{u_2} - \frac{5}{u_2} = -\frac{4}{u_2} \implies u_2 = 4f_0

.

The second object position gives a magnification of

\frac{1}{5}

.

4

Use the known object shift distance to solve for

f_0

and radius of curvature

R

.

u_2 - u_1 = 20\text{ cm} \implies 4f_0 - 2f_0 = 20\text{ cm} \implies 2f_0 = 20\text{ cm} \implies f_0 = 10\text{ cm}

. Since

R = 2f_0

,

R = 20\text{ cm}

.

The distance between the two object positions is

20\text{ cm}

, and the radius of curvature of a spherical mirror is twice its focal length.

Anahtar Kavram

Spherical Mirror Formula and Sign Conventions for Convex Mirrors

Soru 8Soru →

A ray of light strikes a plane mirror at an angle of incidence of $35^\circ$ . What is the angle of deviation of the reflected ray?

110^\circ

70^\circ

55^\circ

145^\circ

Cevabı ve açıklamayı göster

Cevap: $110^\circ$

Cevap

The angle of deviation of the reflected ray is

110^\circ

.

The angle of deviation

d

represents the angle through which a ray of light is turned from its original path. For a plane mirror,

d = 180^\circ - 2i

. Substituting

i = 35^\circ

yields

d = 180^\circ - 70^\circ = 110^\circ

.

Adım Adım Çözüm

1

Identify the given angle of incidence

i = 35^\circ

The angle of incidence is measured relative to the normal line.

2

Apply the law of reflection

Angle of reflection

r = i = 35^\circ

The angle of reflection equals the angle of incidence.

3

Calculate the angle of deviation

d = 180^\circ - (i + r) = 180^\circ - 2(35^\circ) = 110^\circ

The angle of deviation measures how much the light ray is turned from its original initial straight path.

Anahtar Kavram

Angle of deviation for reflection at a plane surface

Tahmini Süre:45s

Soru 9Soru →

A concave mirror forms a real image that is twice the size of an object. When the object is shifted $10\text{ cm}$ closer to the mirror, a virtual image of the same magnification is produced. What is the focal length of the mirror?

10\text{ cm}

15\text{ cm}

20\text{ cm}

30\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $10\text{ cm}$

Cevap

The focal length of the concave mirror is

10\text{ cm}

.

For a concave mirror forming a real image of magnification

2

,

v_1 = 2u_1

, yielding

u_1 = 1.5f

. When the object is moved

10\text{ cm}

closer, a virtual image of magnification

2

is formed, so

v_2 = -2u_2

, yielding

u_2 = 0.5f

. Subtracting the two object positions (

1.5f - 0.5f = 10\text{ cm}

) directly gives

f = 10\text{ cm}

.

Adım Adım Çözüm

1

Set up the mirror equation for the first case (real image).

u_1 = \frac{3}{2}f

For a real inverted image with magnification

m = 2

,

v_1 = +2u_1

. Substituting into

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{v_1}

gives

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{2u_1} = \frac{3}{2u_1}

.

2

Set up the mirror equation for the second case (virtual image).

u_2 = \frac{1}{2}f

For a virtual erect image with magnification

m = 2

, sign convention dictates

v_2 = -2u_2

. Substituting into

\frac{1}{f} = \frac{1}{u_2} + \frac{1}{v_2}

gives

\frac{1}{f} = \frac{1}{u_2} - \frac{1}{2u_2} = \frac{1}{2u_2}

.

3

Use the given displacement between the two object positions to solve for

f

.

f = 10\text{ cm}

The object is moved

10\text{ cm}

closer, so

u_1 - u_2 = 10\text{ cm}

. Substituting the expressions yields

\frac{3}{2}f - \frac{1}{2}f = 10\text{ cm} \implies f = 10\text{ cm}

.

Anahtar Kavram

Mirror Formula and Sign Convention for Spherical Mirrors

Tahmini Süre:2m 0s

Soru 10Soru →

A side-view convex mirror on a bus has a radius of curvature of $40\text{ cm}$ . If a motorcycle is located $30\text{ cm}$ in front of the mirror, what is the location of the image formed relative to the mirror?

12\text{ cm}

behind the mirror

12\text{ cm}

in front of the mirror

60\text{ cm}

behind the mirror

60\text{ cm}

in front of the mirror

Cevabı ve açıklamayı göster

Cevap: $12\text{ cm}$ behind the mirror

Cevap

The image is formed

12\text{ cm}

behind the mirror.

For a convex mirror, the focal length is virtual, so

f = -20\text{ cm}

. With an object distance of

u = +30\text{ cm}

, applying the mirror equation

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

gives

\frac{1}{v} = -\frac{1}{20} - \frac{1}{30} = -\frac{1}{12}

, leading to

v = -12\text{ cm}

. The negative sign specifies that the virtual image is located

12\text{ cm}

behind the mirror.

Adım Adım Çözüm

1

Determine the focal length of the mirror from its radius of curvature

f = -\frac{R}{2} = -\frac{40\text{ cm}}{2} = -20\text{ cm}

For spherical mirrors, focal length is half the radius of curvature. Convex mirrors have a negative focal length by sign convention.

2

Set up the mirror formula using the given object distance

u = +30\text{ cm}

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies -\frac{1}{20} = \frac{1}{30} + \frac{1}{v}

The mirror equation relates focal length, object distance, and image distance.

3

Solve for the image distance

v

\frac{1}{v} = -\frac{1}{20} - \frac{1}{30} = -\frac{3 + 2}{60} = -\frac{5}{60} = -\frac{1}{12} \implies v = -12\text{ cm}

Algebraic manipulation yields a negative image distance.

4

Interpret the physical meaning of the calculated value

The negative sign indicates a virtual image located

12\text{ cm}

behind the mirror.

Under standard optical sign conventions, negative image distances correspond to virtual images formed behind the mirror.

Anahtar Kavram

Mirror equation and sign conventions for convex spherical mirrors

Soru 11Soru →

A convex spherical mirror produces an image that is one-third the size of an object placed in front of it. If the distance of the object from the mirror is doubled, what is the new linear magnification of the image?

\frac{1}{5}

\frac{1}{4}

\frac{1}{6}

\frac{1}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{5}$

Cevap

The new linear magnification of the image is

\frac{1}{5}

.

For a convex mirror, the linear magnification

m

relates object distance

u

and focal magnitude

f

by

m = \frac{f}{f + u}

. Given

m = \frac{1}{3}

, solving

\frac{1}{3} = \frac{f}{f + u}

yields

u = 2f

. When the object distance is doubled to

u' = 4f

, the new magnification becomes

m' = \frac{f}{f + 4f} = \frac{1}{5}

.

Adım Adım Çözüm

1

Express linear magnification in terms of object distance and focal length for a convex mirror

m = \frac{f}{f + u}

For a convex mirror, the focal length is negative under the Cartesian sign convention, making the virtual image distance

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

, so magnification

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

.

2

Substitute the initial magnification

m = \frac{1}{3}

to express initial object distance

u

in terms of focal length

f

\frac{1}{3} = \frac{f}{f + u} \implies f + u = 3f \implies u = 2f

This establishes that the object was originally located at a distance equal to twice the focal length of the mirror.

3

Calculate the new object distance

u'

when distance is doubled

u' = 2u = 2(2f) = 4f

The problem states the object distance from the mirror is doubled.

4

Compute the new linear magnification

m'

m' = \frac{f}{f + u'} = \frac{f}{f + 4f} = \frac{f}{5f} = \frac{1}{5}

Substituting

u' = 4f

into the magnification formula yields the final reduced magnification.

Anahtar Kavram

Linear magnification and sign convention for convex mirrors

Tahmini Süre:2m 0s

Soru 12Soru →

An object is placed at a distance of $15\text{ cm}$ in front of a concave mirror with a focal length of $10\text{ cm}$ . What is the distance of the image formed from the mirror?

30\text{ cm}

6\text{ cm}

25\text{ cm}

-30\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $30\text{ cm}$

Cevap

The distance of the image formed from the mirror is

30\text{ cm}

.

Applying the mirror formula

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

with

f = 10\text{ cm}

and

u = 15\text{ cm}

gives

\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}

, yielding an image distance of

30\text{ cm}

.

Adım Adım Çözüm

1

Identify given quantities and signs

Focal length

f = 10\text{ cm}

and object distance

u = 15\text{ cm}

.

For a concave mirror, the real focus and real object distances are both positive.

2

Set up the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{10} = \frac{1}{15} + \frac{1}{v}

The mirror formula relates object distance, image distance, and focal length.

3

Solve for the image distance

v

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

Subtracting the reciprocals and taking the inverse gives the image position.

Anahtar Kavram

Concave Mirror Formula

Soru 13Soru →

A concave shaving mirror has a radius of curvature of $60\text{ cm}$ . A person places their face in front of the mirror such that an upright image magnified $3$ times is formed. What is the distance of the face from the mirror, in centimeters?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The distance of the person's face from the mirror is

20\text{ cm}

.

For a concave mirror with a radius of curvature of

60\text{ cm}

, the focal length is

f = +30\text{ cm}

. An upright image is virtual, corresponding to a positive magnification

m = +3

. Since

m = -\frac{v}{u}

, the image distance is

v = -3u

. Substituting these into the mirror formula

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

gives

\frac{1}{30} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

, solving to

u = 20\text{ cm}

.

Adım Adım Çözüm

1

Determine the focal length of the concave mirror.

f = 30\text{ cm}

The focal length is half the radius of curvature (

f = \frac{r}{2} = \frac{60\text{ cm}}{2} = 30\text{ cm}

).

2

Express the image distance

v

in terms of the object distance

u

using the magnification relationship.

v = -3u

An upright image produced by a spherical mirror is virtual, so linear magnification

m = +3

. Using

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

, we obtain

v = -3u

.

3

Substitute

f

and

v

into the mirror equation to solve for

u

.

u = 20\text{ cm}

Applying the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{30} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u} \implies 3u = 60 \implies u = 20\text{ cm}

.

Anahtar Kavram

Mirror equation and sign conventions for virtual images formed by concave mirrors

Soru 14Soru →

An object is placed at a distance $u$ in front of a concave mirror of focal length $12\text{ cm}$ . A plane mirror is placed perpendicular to the principal axis at a distance of $32\text{ cm}$ in front of the concave mirror, between the object and the concave mirror. If the real image formed by the concave mirror coincides in space with the virtual image formed by the plane mirror, what is the value of $u$ in centimeters?

Cevabı ve açıklamayı göster

Cevap: 48

Cevap

The correct object distance

u

is

48\text{ cm}

.

The object is located at distance

u

from the concave mirror. With the plane mirror at

32\text{ cm}

from the concave mirror, the object distance from the plane mirror is

u - 32

. The plane mirror forms an image at distance

u - 32

behind itself, which corresponds to

32 - (u - 32) = 64 - u

from the concave mirror. Setting

v = 64 - u

in the mirror formula

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

gives

u^2 - 64u + 768 = 0

. Factoring yields

u = 48\text{ cm}

or

u = 16\text{ cm}

. Because the plane mirror is between the object and the concave mirror,

u > 32\text{ cm}

, so

u = 48\text{ cm}

.

Adım Adım Çözüm

1

Find the position of the image formed by the plane mirror in terms of

u

.

The object is at a distance

(u - 32)\text{ cm}

in front of the plane mirror. Its virtual image is formed

(u - 32)\text{ cm}

behind the plane mirror, which places it at

32 - (u - 32) = (64 - u)\text{ cm}

in front of the concave mirror.

A plane mirror forms an image behind it at a distance equal to the object distance in front of it.

2

Equate the image distance of the concave mirror

v

to the position of the plane mirror image.

v = 64 - u

The question states that the image formed by the concave mirror coincides in position with the image formed by the plane mirror.

3

Substitute

f = 12\text{ cm}

and

v = 64 - u

into the mirror equation.

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

The standard mirror formula relates focal length, object distance, and image distance.

4

Solve the algebraic equation for

u

.

\frac{1}{12} = \frac{(64 - u) + u}{u(64 - u)} \implies 64u - u^2 = 768 \implies u^2 - 64u + 768 = 0

Combining fractions and multiplying across gives a quadratic equation in standard form.

5

Factor the quadratic equation and select the physical root.

(u - 48)(u - 16) = 0 \implies u = 48\text{ cm}

or

u = 16\text{ cm}

. Since

u > 32\text{ cm}

,

u = 48\text{ cm}

.

The plane mirror is situated between the object and the concave mirror at

32\text{ cm}

, so the object distance

u

must be greater than

32\text{ cm}

.

Anahtar Kavram

Image coincidence in combined plane and curved optical systems

Tahmini Süre:3m 0s

Soru 15Soru →

An object is placed $24\text{ cm}$ in front of a concave mirror with a focal length of $8\text{ cm}$ . What is the distance of the image from the mirror in $\text{cm}$ ?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The distance of the image from the mirror is

12\text{ cm}

.

Applying the spherical mirror formula

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

with focal length

f = 8\text{ cm}

and object distance

u = 24\text{ cm}

yields

\frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{1}{12}

, giving an image distance of

12\text{ cm}

.

Adım Adım Çözüm

1

Identify the given parameters and select the appropriate relation.

Focal length

f = 8\text{ cm}

, object distance

u = 24\text{ cm}

, using the mirror equation

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

.

The mirror formula relates focal length, object distance, and image distance for spherical mirrors.

2

Substitute the given values into the formula and solve for

\frac{1}{v}

.

\frac{1}{8} = \frac{1}{24} + \frac{1}{v} \implies \frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{3-1}{24} = \frac{2}{24} = \frac{1}{12}

Subtracting

\frac{1}{24}

from

\frac{1}{8}

isolates the reciprocal of the image distance.

3

Invert the result to determine the image distance

v

.

v = 12\text{ cm}

Taking the reciprocal of

\frac{1}{12}

gives the image distance in centimeters.

Anahtar Kavram

Mirror Formula for Concave Mirrors

Tahmini Süre:45s

Soru 16Soru →

A concave mirror produces a real image that is $3$ times the size of an object placed in front of it. If the distance between the object and its image is $40\text{ cm}$ , what is the focal length of the mirror in $\text{cm}$ ?

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

The focal length of the concave mirror is

15\text{ cm}

.

Using the magnification relation

v = 3u

and the object-image separation of

40\text{ cm}

, we obtain

3u - u = 40\text{ cm}

, which yields

u = 20\text{ cm}

and

v = 60\text{ cm}

. Substituting these distances into the mirror equation

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

gives

\frac{1}{f} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

, so

f = 15\text{ cm}

.

Adım Adım Çözüm

1

Relate the image distance

v

to the object distance

u

using the linear magnification formula

v = 3u

Since the mirror forms a real, magnified image that is 3 times the size of the object, linear magnification

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

.

2

Formulate an equation from the given object-to-image separation distance to solve for

u

and

v

u = 20\text{ cm}

and

v = 60\text{ cm}

The separation distance between the image and object is

v - u = 40\text{ cm}

. Substituting

v = 3u

yields

2u = 40\text{ cm} \implies u = 20\text{ cm}

and

v = 60\text{ cm}

.

3

Substitute the values of

u

and

v

into the mirror formula to compute the focal length

f

f = 15\text{ cm}

Applying

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

gives

f = 15\text{ cm}

.

Anahtar Kavram

Linear magnification and mirror formula for concave mirrors

Soru 17Soru →

A light ray strikes a plane mirror. Keeping the direction of the incident ray fixed, the mirror is rotated through an angle of $15^\circ$ about an axis lying in its plane. What is the angle of rotation, in degrees, of the reflected ray?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The reflected ray turns through an angle of

30^\circ

.

When a plane mirror is rotated through an angle

\theta

while keeping the incident ray direction constant, the normal turns by

\theta

. This changes the angle of incidence by

\theta

and the angle of reflection by

\theta

, causing the reflected ray to rotate by a total angle of

2\theta

. For a mirror rotation of

15^\circ

, the reflected ray rotates through

2 \times 15^\circ = 30^\circ

.

Adım Adım Çözüm

1

Analyze the effect of mirror rotation on the normal line

When the plane mirror rotates by

15^\circ

, the normal to the mirror surface also rotates by

15^\circ

.

The normal line is always perpendicular to the surface of the mirror.

2

Determine the change in the angle of incidence and reflection

The angle of incidence changes by

15^\circ

, so the angle of reflection relative to the new normal also changes by

15^\circ

.

According to the law of reflection, the angle of incidence equals the angle of reflection (

i = r

).

3

Calculate the total angular deviation of the reflected ray

The total shift of the reflected ray relative to its original path is

15^\circ + 15^\circ = 30^\circ

.

The rotation of the reflected ray is twice the angle of rotation of the mirror.

Anahtar Kavram

Rotation of Reflected Ray by a Plane Mirror

Soru 18Soru →

A convex mirror used as a security mirror in a store has a radius of curvature of $20\text{ cm}$ . If a shopper stands $30\text{ cm}$ in front of the mirror, at what distance from the mirror is the shopper's image formed?

7.5\text{ cm}

behind the mirror

15.0\text{ cm}

in front of the mirror

12.0\text{ cm}

behind the mirror

15.0\text{ cm}

behind the mirror

Cevabı ve açıklamayı göster

Cevap: $7.5\text{ cm}$ behind the mirror

Cevap

The shopper's image is formed

7.5\text{ cm}

behind the mirror.

For a convex mirror, the focal length is negative and given by

f = -r/2 = -10\text{ cm}

. Substituting

f = -10\text{ cm}

and object distance

u = +30\text{ cm}

into the mirror equation

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

gives

\frac{1}{v} = -\frac{1}{10} - \frac{1}{30} = -\frac{4}{30}

, which solves to

v = -7.5\text{ cm}

. A negative image distance represents a virtual image formed

7.5\text{ cm}

behind the mirror.

Adım Adım Çözüm

1

Determine the focal length of the convex mirror using its radius of curvature

f = -\frac{r}{2} = -\frac{20\text{ cm}}{2} = -10\text{ cm}

By sign convention, a convex mirror has a negative focal length equal to half its radius of curvature.

2

Apply the mirror formula to find the image distance

v

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{-10} = \frac{1}{30} + \frac{1}{v}

The mirror formula relates focal length (

f

), object distance (

u

), and image distance (

v

).

3

Solve for

\frac{1}{v}

and calculate

v

\frac{1}{v} = -\frac{1}{10} - \frac{1}{30} = -\frac{3 + 1}{30} = -\frac{4}{30} = -\frac{2}{15}\text{ cm}^{-1} \implies v = -7.5\text{ cm}

The negative sign indicates that the image is virtual and located behind the mirror.

Anahtar Kavram

Mirror Formula and Sign Convention for Convex Mirrors

Tahmini Süre:1m 30s

Soru 19Soru →

A dentist uses a concave mirror to examine a patient's tooth. When the mirror is placed $12\text{ cm}$ in front of the tooth, it forms an erect image that is magnified $3$ times. What is the focal length of the mirror?

18\text{ cm}

9\text{ cm}

24\text{ cm}

36\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $18\text{ cm}$

Cevap

The focal length of the mirror is

18\text{ cm}

.

For a concave mirror, an erect image is always virtual, located behind the mirror. The magnification formula

m = -v/u = +3

gives an image distance of

v = -36\text{ cm}

when the object distance

u = 12\text{ cm}

. Applying the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{12} - \frac{1}{36} = \frac{2}{36} = \frac{1}{18}

yields a focal length of

18\text{ cm}

.

Adım Adım Çözüm

1

Identify the nature of the image and apply the magnification relationship

Since the concave mirror forms an erect image, the image must be virtual. Therefore, linear magnification

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

.

An erect image formed by a spherical mirror is always virtual, which corresponds to a negative image distance under standard sign conventions.

2

Calculate the image distance

v

v = -3 \times 12\text{ cm} = -36\text{ cm}

.

Given object distance

u = +12\text{ cm}

, multiplying by

-3

yields the position of the virtual image behind the mirror.

3

Substitute

u

and

v

into the mirror formula to find

f

\frac{1}{f} = \frac{1}{12} + \frac{1}{-36} = \frac{3 - 1}{36} = \frac{2}{36} = \frac{1}{18} \implies f = +18\text{ cm}.

The mirror equation

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

relates object distance, image distance, and focal length.

Anahtar Kavram

Focal length calculation for spherical mirrors producing virtual images

Soru 20Soru →

A convex spherical mirror always forms a virtual, erect, and diminished image of a real object, regardless of the object's distance from the mirror.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is True. A convex mirror consistently produces virtual, erect, and diminished images for all real object positions.

The statement is correct because the outward curvature of a convex mirror causes all incident parallel or diverging rays from a real object to diverge upon reflection. The virtual extensions of these rays converge behind the mirror between the pole and the focus, ensuring the image is always virtual, upright, and smaller than the object.

Adım Adım Çözüm

1

Identify the sign conventions for a convex mirror and a real object.

The focal length

f

is negative (

f < 0

) because the focus is behind the mirror, and the object distance

u

is positive (

u > 0

) for a real object.

Establishing proper sign convention is essential for analyzing image formation in curved mirrors.

2

Analyze the mirror equation to determine the sign of the image distance

v

.

From

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

, re-arranging gives

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\left(\frac{1}{|f|} + \frac{1}{u}\right)

. Thus,

v

is always negative.

A negative image distance (

v < 0

) mathematically proves that the image is virtual and located behind the mirror.

3

Evaluate linear magnification

m

to determine image orientation and size.

Using

m = -\frac{v}{u}

, since

v < 0

and

u > 0

,

m > 0

(erect image). Additionally,

|v| = \frac{|f|u}{u + |f|} < u

, so

|m| = \frac{|v|}{u} < 1

(diminished image).

Magnification sign indicates orientation (positive is erect) and magnitude indicates size relative to the object.

Anahtar Kavram

Image characteristics in convex mirrors

Reflection of Light at Plane and Curved Mirrors

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