Wave Phenomena: Reflection, Refraction, Interference, Diffraction, and Polarization Practice Questions — JAMB UTME

Question 1Question →

Monochromatic light of wavelength $500\text{ nm}$ is incident normally on a plane diffraction grating. If the second-order principal maximum is observed at an angle of $30^\circ$ to the normal, calculate the number of lines per millimeter ruled on the grating.

Show answer & explanation

Answer: 500

Answer

500 lines/mm

Using the diffraction grating equation

d \sin\theta = n\lambda

with

n = 2

,

\lambda = 5.0 \times 10^{-7}\text{ m}

, and

\sin(30^\circ) = 0.5

gives a slit separation of

d = 2.0 \times 10^{-6}\text{ m}

. Converting to grating ruling density per millimeter yields

\frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}

.

Step-by-Step Solution

1

Identify the given physical parameters and convert wavelength to standard SI meters.

Order

n = 2

, diffraction angle

\theta = 30^\circ

, wavelength

\lambda = 500 \times 10^{-9}\text{ m} = 5.0 \times 10^{-7}\text{ m}

.

Standard SI unit conversion is required to perform wave equation calculations accurately.

2

Apply the diffraction grating equation

d \sin\theta = n\lambda

to compute the grating spacing

d

.

d = \frac{2 \times 5.0 \times 10^{-7}\text{ m}}{\sin 30^\circ} = \frac{1.0 \times 10^{-6}}{0.5} = 2.0 \times 10^{-6}\text{ m}

.

The condition for principal constructive interference maxima is

d \sin\theta = n\lambda

.

3

Calculate the number of lines per millimeter by dividing

1\text{ mm}

(

10^{-3}\text{ m}

) by the grating spacing

d

.

N_{\text{mm}} = \frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}

.

Grating density (lines per unit length) is the reciprocal of the slit separation

d

.

Key Concept

Diffraction Grating Principal Maxima Condition

Estimated Time:2m 0s

Question 2Question →

A light wave of frequency $5.00 \times 10^{14}\text{ Hz}$ travels from air into a glass medium with a refractive index of $1.50$ . What is the frequency of the light wave inside the glass medium?

3.33 \times 10^{14}\text{ Hz}

5.00 \times 10^{14}\text{ Hz}

7.50 \times 10^{14}\text{ Hz}

2.25 \times 10^{14}\text{ Hz}

Show answer & explanation

Answer: $5.00 \times 10^{14}\text{ Hz}$

Answer

The frequency of the light wave inside the glass medium remains

5.00 \times 10^{14}\text{ Hz}

.

When light passes from one medium into another, its speed and wavelength change, but its frequency remains constant because it depends solely on the source producing the wave.

Step-by-Step Solution

1

Identify the given physical parameters.

Initial frequency

f = 5.00 \times 10^{14}\text{ Hz}

and refractive index

n = 1.50

.

Understanding provided quantities is the starting point for wave refraction analysis.

2

Apply the principle of frequency conservation across optical boundaries.

Frequency in glass medium

f_{\text{glass}} = f_{\text{air}} = 5.00 \times 10^{14}\text{ Hz}

.

When a wave passes into a different medium, its speed and wavelength alter in proportion to the refractive index, but its frequency is determined solely by the periodic source and remains constant.

Key Concept

Invariance of Wave Frequency across Refracting Media

Question 3Question →

In a Young's double-slit experiment, the separation between two narrow slits is $0.40\text{ mm}$ and the interference pattern is observed on a screen placed $1.20\text{ m}$ away from the slits. If the distance between consecutive bright fringes on the screen is $1.80\text{ mm}$ , what is the wavelength of the light used in nanometers ( $\text{nm}$ )?

Show answer & explanation

Answer: 600

Answer

The wavelength of the light used is

600\text{ nm}

.

Using the Young's double-slit fringe spacing relation

\beta = \frac{\lambda D}{d}

, rearranging yields

\lambda = \frac{\beta d}{D}

. Substituting

\beta = 1.80 \times 10^{-3}\text{ m}

,

d = 4.0 \times 10^{-4}\text{ m}

, and

D = 1.20\text{ m}

gives

\lambda = 6.00 \times 10^{-7}\text{ m}

, which corresponds to

600\text{ nm}

.

Step-by-Step Solution

1

Convert given physical quantities into standard SI units (meters).

Slit separation

d = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}

, distance to screen

D = 1.20\text{ m}

, and fringe spacing

\beta = 1.80\text{ mm} = 1.80 \times 10^{-3}\text{ m}

.

Standard SI units ensure accuracy when applying wave speed and distance equations.

2

Write the Young's double-slit formula relating fringe width to wavelength.

\beta = \frac{\lambda D}{d}

This relationship defines the spatial period of interference fringes on a screen.

3

Rearrange the equation to isolate the wavelength

\lambda

.

\lambda = \frac{\beta d}{D}

The unknown parameter to solve for is the wavelength of the monochromatic source.

4

Substitute the numerical values and convert the final result to nanometers.

$\lambda = \frac{1.80 \times 10^{-3}\text{ m} \times 4.0 \times 10^{-4}\text{ m}}{1.20\text{ m}} = 6.00 \times 10^{-7}\text{ m} = 600\text{ nm}$

Multiply meters by

10^9

to express the wavelength in nanometers.

Key Concept

Young's Double-Slit Interference Fringe Spacing

Question 4Question →

A light ray travels from Medium X into Medium Y. The speed of light in Medium X is $1.50 \times 10^8\text{ m/s}$ and the speed of light in Medium Y is $2.50 \times 10^8\text{ m/s}$ . What is the sine of the critical angle for total internal reflection at the boundary between these two media?

0.60

1.67

0.50

0.40

Show answer & explanation

Answer: $0.60$

Answer

The sine of the critical angle for total internal reflection at the boundary is

0.60

.

Total internal reflection occurs when light moves from an optically denser medium to an optically less dense medium. By Snell's law,

n_X \sin\theta_c = n_Y \sin 90^\circ

, which gives

\sin\theta_c = n_Y / n_X

. Since refractive index is inversely proportional to speed (

n = c/v

),

\sin\theta_c = v_X / v_Y = (1.50 \times 10^8) / (2.50 \times 10^8) = 0.60

.

Step-by-Step Solution

1

Relate speed of light in each medium to their respective refractive indices

n_X = \frac{c}{v_X}

and

n_Y = \frac{c}{v_Y}

Refractive index is defined as the ratio of light speed in vacuum to light speed in the medium.

2

Apply Snell's Law for the critical angle condition

\sin\theta_c = \frac{n_Y}{n_X}

At the critical angle, the angle of refraction in the less dense medium is

90^\circ

(

\sin 90^\circ = 1

).

3

Substitute light speeds into the critical angle formula and calculate

\sin\theta_c = \frac{v_X}{v_Y} = \frac{1.50 \times 10^8\text{ m/s}}{2.50 \times 10^8\text{ m/s}} = 0.60

Substituting the expressions for

n_X

and

n_Y

simplifies the refractive index ratio to the direct ratio of speeds

v_X / v_Y

.

Key Concept

Critical angle and total internal reflection relation to speed of light in media

Question 5Question →

Monochromatic light of wavelength $600\text{ nm}$ is incident normally on a diffraction grating having $500\text{ lines per mm}$ . What is the angle of diffraction, in degrees, for the first-order principal maximum?

Show answer & explanation

Answer: 17.5

Answer

The angle of diffraction for the first-order principal maximum is

17.5^\circ

.

Using the grating equation

d \sin \theta = n \lambda

, the slit separation is

d = \frac{10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}

. For

n = 1

and

\lambda = 6.00 \times 10^{-7}\text{ m}

, we get

\sin \theta = \frac{6.00 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.30

. Taking

\arcsin(0.30)

gives approximately

17.5^\circ

.

Step-by-Step Solution

1

Calculate the grating element (slit spacing)

d

d = 2.00 \times 10^{-6}\text{ m}

Grating spacing

d

is the reciprocal of the line density

N = 500\text{ lines/mm} = 500,000\text{ lines/m}

.

2

Apply the diffraction grating equation

d \sin \theta = n \lambda

\sin \theta = 0.30

For the first-order maximum (

n = 1

),

\sin \theta = \frac{1 \times 600 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.30

.

3

Find the angle

\theta

by taking the inverse sine

\theta = 17.5^\circ

\arcsin(0.30) \approx 17.46^\circ

, which rounds to

17.5^\circ

.

Key Concept

Diffraction Grating Equation for Principal Maxima

Estimated Time:1m 30s

Question 6Question →

Two sound waves of frequencies $440\text{ Hz}$ and $445\text{ Hz}$ travel through air and superpose to produce beats. What is the resulting beat frequency, in hertz?

Show answer & explanation

Answer: 5

Answer

The beat frequency produced by the superposition of the two sound waves is

5\text{ Hz}

.

When two waves of slightly different frequencies interfere, periodic variations in sound intensity called beats occur. The number of beats heard per second is equal to the absolute difference between the frequencies of the two superposing waves:

f_{\text{beat}} = |f_2 - f_1| = |445\text{ Hz} - 440\text{ Hz}| = 5\text{ Hz}

.

Step-by-Step Solution

1

Identify the frequencies of the interfering waves

f_1 = 440\text{ Hz}

and

f_2 = 445\text{ Hz}

Beat frequency is determined by the absolute difference between the individual wave frequencies.

2

Subtract the lower frequency from the higher frequency to find the beat frequency

f_{\text{beat}} = |445 - 440| = 5\text{ Hz}

The rate of periodic intensity variation (beat frequency) is governed by

f_{\text{beat}} = |f_2 - f_1|

.

Key Concept

Beat Frequency and Wave Superposition

Question 7Question →

A monochromatic beam of light with a wavelength of $600\text{ nm}$ in air enters a glass block of refractive index $1.50$ . What is the wavelength of the light inside the glass block in nanometers ( $\text{nm}$ )?

Show answer & explanation

Answer: 400

Answer

The wavelength of the light inside the glass block is

400\text{ nm}

.

The speed and wavelength of light both decrease by a factor of the refractive index

n

when entering a medium from air, while the frequency stays constant (

v = f\lambda

). Thus,

\lambda = \frac{\lambda_0}{n} = \frac{600}{1.50} = 400\text{ nm}

.

Step-by-Step Solution

1

Identify the relevant formula connecting refractive index and wavelength

\lambda = \frac{\lambda_0}{n}

When light passes from air into a medium, its frequency remains unchanged while its speed and wavelength decrease proportionally by a factor of the refractive index

n

.

2

Substitute the given values into the equation

\lambda = \frac{600\text{ nm}}{1.50}

The wavelength in air is

600\text{ nm}

and the refractive index of glass is

1.50

.

3

Perform the calculation

\lambda = 400\text{ nm}

Dividing

600

by

1.50

yields

400

.

Key Concept

Refraction and Wavelength Change in a Medium

Question 8Question →

Under which of the following conditions will total internal reflection occur when light encounters the boundary between two transparent media?

When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence greater than the critical angle.

When light travels from a medium of lower refractive index to one of higher refractive index at an angle of incidence greater than the critical angle.

When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence less than the critical angle.

When light travels from a medium of lower refractive index to one of higher refractive index at normal incidence.

Show answer & explanation

Answer: When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence greater than the critical angle.

Answer

Total internal reflection occurs when light travels from a medium of higher refractive index to a medium of lower refractive index at an angle of incidence greater than the critical angle.

Total internal reflection takes place only when light moves from an optically denser medium (higher refractive index) into an optically less dense medium (lower refractive index), and the angle of incidence at the interface is greater than the critical angle for the two media.

Step-by-Step Solution

1

Identify the optical density requirement for total internal reflection

Light must travel from an optically denser medium (higher refractive index

n_1

) toward an optically rarer medium (lower refractive index

n_2

).

This condition ensures that the light refracts away from the normal into the second medium.

2

Identify the angular requirement at the interface

The angle of incidence

i

must be strictly greater than the critical angle

\theta_c

(

\sin\theta_c = \frac{n_2}{n_1}

).

When

i > \theta_c

, no light can refract into the second medium, resulting in total reflection back into the initial medium.

Key Concept

Conditions for Total Internal Reflection

Question 9Question →

A ray of light traveling inside a transparent medium strikes the boundary with air. If the critical angle for total internal reflection at this boundary is $30^\circ$ , what is the refractive index of the medium?

2.00

0.50

0.87

1.15

Show answer & explanation

Answer: $2.00$

Answer

The refractive index of the medium is

2.00

.

For light traveling from a medium into air, the critical angle

C

is related to the refractive index

n

by

n = \frac{1}{\sin C}

. Substituting

C = 30^\circ

gives

n = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2.00

.

Step-by-Step Solution

1

Identify the relationship between critical angle

C

and refractive index

n

The formula for light passing into air is

n = \frac{1}{\sin C}

.

Total internal reflection occurs when light travels from a denser medium to a less dense medium (air,

n_{air} = 1

) at an angle greater than the critical angle.

2

Substitute the given critical angle

C = 30^\circ

into the formula

\sin 30^\circ = 0.5

, so

n = \frac{1}{0.5} = 2.00

.

Taking the reciprocal of

\sin 30^\circ

gives the correct refractive index.

Key Concept

Critical Angle and Refractive Index

Question 10Question →

In a double-slit experiment, monochromatic light of wavelength $500\text{ nm}$ is incident normally on two narrow slits. If the third-order ( $m = 3$ ) bright fringe is observed at an angle of $30^\circ$ from the central maximum, what is the slit separation, $d$ , in micrometers ( $\mu\text{m}$ )?

Show answer & explanation

Answer: 3

Answer

The slit separation is

3.0\text{ }\mu\text{m}

.

For bright fringes in a double-slit setup, constructive interference occurs when

d \sin\theta = m\lambda

. Rearranging for slit separation yields

d = \frac{m\lambda}{\sin\theta}

. Substituting

m = 3

,

\lambda = 0.50\text{ }\mu\text{m}

, and

\theta = 30^\circ

gives

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{\sin 30^\circ} = \frac{1.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

.

Step-by-Step Solution

1

Identify the constructive interference equation for Young's double-slit experiment.

d \sin\theta = m\lambda

Bright fringes occur where waves from the two slits interfere constructively, which corresponds to path differences equal to integer multiples of the wavelength.

2

Convert the given wavelength to micrometers and substitute known values.

\lambda = 0.50\text{ }\mu\text{m}

,

m = 3

,

\sin(30^\circ) = 0.50

Converting units to micrometers early simplifies direct calculation of

d

in

\mu\text{m}

.

3

Rearrange the equation and evaluate for

d

.

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

Dividing the numerator by

\sin(30^\circ) = 0.50

doubles the value of

m\lambda

.

Key Concept

Angular position condition for constructive interference in double-slit diffraction

Question 11Question →

An unpolarized light beam with an initial intensity of $80\text{ W/m}^2$ passes through an ideal linear polarizer. What is the intensity of the transmitted light in $\text{W/m}^2$ ?

Show answer & explanation

Answer: 40

Answer

The intensity of the transmitted light is

40\text{ W/m}^2

.

When unpolarized light of initial intensity

I_0

encounters an ideal linear polarizing filter, the transmitted intensity

I

is always equal to half of the incident intensity (

I = \frac{1}{2}I_0

). Substituting

80\text{ W/m}^2

gives

I = 40\text{ W/m}^2

.

Step-by-Step Solution

1

Determine the fraction of unpolarized light intensity transmitted by a polarizer.

Transmitted intensity formula

I = \frac{1}{2} I_0

.

Unpolarized light consists of randomly oriented electric field vectors, resulting in an average transmission factor of one-half.

2

Substitute the incident intensity value into the formula.

I = \frac{80}{2} = 40\text{ W/m}^2

.

Direct mathematical calculation.

Key Concept

Polarization and intensity reduction of unpolarized light upon passing through a linear polarizer.

Question 12Question →

A beam of light traveling in air is incident on a transparent liquid at the polarizing angle (Brewster's angle) of $53.1^\circ$ , where $\tan 53.1^\circ = 1.33$ . What is the critical angle for total internal reflection when light travels from this liquid into air?

\sin^{-1}(0.75)

\sin^{-1}(0.60)

\sin^{-1}(1.33)

53.1^\circ

Show answer & explanation

Answer: $\sin^{-1}(0.75)$

Answer

The critical angle for total internal reflection at the liquid-air boundary is

\sin^{-1}(0.75)

.

According to Brewster's law, the refractive index of the liquid is given by

n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}

. When light travels from the denser liquid medium to the rarer air medium, the critical angle

\theta_c

for total internal reflection satisfies

\sin\theta_c = \frac{1}{n}

. Substituting

n = \frac{4}{3}

gives

\sin\theta_c = \frac{3}{4} = 0.75

, so

\theta_c = \sin^{-1}(0.75)

.

Step-by-Step Solution

1

Determine the refractive index of the liquid using Brewster's law.

n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}

.

Brewster's law states that when light in air (

n_1 = 1

) is incident at the polarizing angle

\theta_B

on a medium of index

n

,

\tan\theta_B = n

.

2

Apply the total internal reflection condition for light passing from liquid to air.

\sin\theta_c = \frac{1}{n} = \frac{1}{4/3} = \frac{3}{4} = 0.75

.

Total internal reflection occurs at an interface when light travels from a denser medium (

n

) to a less dense medium (

1

) at an angle greater than

\theta_c

, where

\sin\theta_c = \frac{1}{n}

.

3

Solve for the critical angle

\theta_c

.

\theta_c = \sin^{-1}(0.75)

.

Taking the inverse sine of

0.75

yields the critical angle.

Key Concept

Synthesizing Brewster's law of polarization with total internal reflection critical angle

Question 13Question →

A ray of light travels from air into a liquid with a refractive index of $1.33$ . If the sine of the angle of incidence in air is $0.80$ , what is the sine of the angle of refraction in the liquid?

Show answer & explanation

Answer: 0.6

Answer

The sine of the angle of refraction in the liquid is 0.60.

According to Snell's law for light passing from air into a medium, the refractive index

n

is given by

n = \frac{\sin i}{\sin r}

. Rearranging this equation to solve for the sine of the angle of refraction yields

\sin r = \frac{\sin i}{n}

. Substituting

\sin i = 0.80

and

n = 1.33

(or

\frac{4}{3}

) gives

\sin r = \frac{0.80}{4/3} = 0.60

.

Step-by-Step Solution

1

Identify the given physical parameters and state Snell's law

Refractive index

n = 1.33

(or

\frac{4}{3}

),

\sin i = 0.80

. Snell's law:

n = \frac{\sin i}{\sin r}

Snell's law relates the ratio of the sines of the angles of incidence and refraction to the refractive index of the medium.

2

Rearrange the equation to express the sine of the angle of refraction

\sin r = \frac{\sin i}{n}

Algebraically isolating

\sin r

allows direct substitution of the known quantities.

3

Substitute the values and compute the result

\sin r = \frac{0.80}{4/3} = 0.60

Dividing

0.80

by

\frac{4}{3}

gives

0.60

.

Key Concept

Snell's Law of Refraction

Alternative Method

Convert decimal numbers into simple fractions:

n = \frac{4}{3}

and

\sin i = \frac{4}{5}

. Evaluating

\sin r = \frac{4/5}{4/3}

simplifies directly to

\frac{3}{5} = 0.60

.

Estimated Time:45s

Question 14Question →

A ray of light traveling within a dense glass prism of refractive index $1.60$ strikes the boundary with a surrounding transparent liquid. If total internal reflection just occurs at an angle of incidence of $45.0^\circ$ in the glass, what is the refractive index of the liquid? (Take $\sin 45.0^\circ = 0.707$ )

1.13

2.26

0.44

1.60

Show answer & explanation

Answer: $1.13$

Answer

The refractive index of the liquid is

1.13

.

For light traveling from a denser medium (

n_{\text{glass}}

) to a rarer medium (

n_{\text{liquid}}

), the critical angle

\theta_c

is defined by

\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}

. Substituting

n_{\text{glass}} = 1.60

and

\sin 45.0^\circ = 0.707

gives

n_{\text{liquid}} = 1.60 \times 0.707 = 1.13

.

Step-by-Step Solution

1

Identify the given parameters and formula for total internal reflection

Refractive index of denser medium

n_{\text{glass}} = 1.60

, critical angle

\theta_c = 45.0^\circ

, and formula

\sin \theta_c = \frac{n_{\text{rarer}}}{n_{\text{denser}}}

.

Total internal reflection occurs at the critical angle when light travels from an optically denser medium to a less dense (rarer) medium.

2

Rearrange the equation to solve for the refractive index of the liquid (

n_{\text{liquid}}

)

n_{\text{liquid}} = n_{\text{glass}} \times \sin \theta_c

.

Multiplying both sides of the critical angle equation by

n_{\text{glass}}

isolates the target variable.

3

Substitute the values and calculate

n_{\text{liquid}}

n_{\text{liquid}} = 1.60 \times 0.707 = 1.1312 \approx 1.13

.

Carrying out the arithmetic yields the refractive index of the liquid.

Key Concept

Total Internal Reflection and Critical Angle

Question 15Question →

A water wave traveling in deep water has a wavelength of $0.80\text{ m}$ and a speed of $2.4\text{ m/s}$ . Upon entering a shallow region, its speed drops to $1.8\text{ m/s}$ . What are the frequency and wavelength of the wave in the shallow region?

Frequency =

3.0\text{ Hz}

, Wavelength =

0.60\text{ m}

Frequency =

2.25\text{ Hz}

, Wavelength =

0.80\text{ m}

Frequency =

2.25\text{ Hz}

, Wavelength =

0.60\text{ m}

Frequency =

0.33\text{ Hz}

, Wavelength =

0.60\text{ m}

Show answer & explanation

Answer: Frequency = $3.0\text{ Hz}$ , Wavelength = $0.60\text{ m}$

Answer

Frequency =

3.0\text{ Hz}

, Wavelength =

0.60\text{ m}

When a wave passes from deep to shallow water (refraction), its frequency remains unchanged because frequency is fixed by the source. Using

v = f\lambda

, the initial frequency is

f = \frac{2.4}{0.80} = 3.0\text{ Hz}

. In shallow water, the new wavelength is

\lambda' = \frac{1.8}{3.0} = 0.60\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in deep water using the wave equation

v = f \lambda

.

f = \frac{v}{\lambda} = \frac{2.4\text{ m/s}}{0.80\text{ m}} = 3.0\text{ Hz}

.

The frequency depends on the wave source and can be determined from the given initial speed and wavelength.

2

Apply the boundary condition for wave refraction.

The frequency in shallow water remains

f = 3.0\text{ Hz}

.

When a wave travels from one medium to another, its frequency remains constant.

3

Calculate the new wavelength in shallow water using

\lambda' = \frac{v'}{f}

.

\lambda' = \frac{1.8\text{ m/s}}{3.0\text{ Hz}} = 0.60\text{ m}

.

The wavelength changes proportionally with speed when frequency is constant.

Key Concept

Constancy of wave frequency during refraction across medium boundaries

Question 16Question →

A beam of monochromatic light of wavelength $6.25 \times 10^{-7}\text{ m}$ is incident normally on a plane diffraction grating having $400\text{ lines/mm}$ . What is the angle of diffraction for the second-order principal maximum?

14.5^\circ

30.0^\circ

45.0^\circ

60.0^\circ

Show answer & explanation

Answer: $30.0^\circ$

Answer

The angle of diffraction for the second-order principal maximum is

30.0^\circ

.

The grating spacing is

d = \frac{1}{400\times 10^3\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}

. Using

d \sin\theta = n\lambda

for

n=2

and

\lambda = 6.25 \times 10^{-7}\text{ m}

gives

\sin\theta = \frac{2 \times 6.25 \times 10^{-7}}{2.50 \times 10^{-6}} = 0.50

, corresponding to

\theta = 30.0^\circ

.

Step-by-Step Solution

1

Calculate the grating spacing (

d

).

d = \frac{1\text{ mm}}{400} = 2.50 \times 10^{-3}\text{ mm} = 2.50 \times 10^{-6}\text{ m}

.

Grating spacing

d

is the reciprocal of the line density

N

per unit length.

2

Apply the diffraction grating equation for the second order (

n=2

).

d \sin\theta = n\lambda \implies (2.50 \times 10^{-6}) \sin\theta = 2 \times (6.25 \times 10^{-7}) = 1.25 \times 10^{-6}

.

The principal maximum condition relates slit spacing, diffraction angle, spectral order, and wavelength.

3

Solve for the diffraction angle

\theta

.

\sin\theta = \frac{1.25 \times 10^{-6}}{2.50 \times 10^{-6}} = 0.50 \implies \theta = \arcsin(0.50) = 30.0^\circ

.

Taking the inverse sine of

0.50

yields the exact angle of diffraction.

Key Concept

Diffraction Grating Equation (

d \sin\theta = n\lambda

)

Wave Phenomena: Reflection, Refraction, Interference, Diffraction, and Polarization

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Alternative Method

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