Wave Optics Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the frequency of light with a wavelength of \( 595 \, \text{nm} \) in air, given the speed of light in air is \( 3.0 \times 10^8 \, \text{m/s} \)?

Frequency \( \nu = \frac{c}{\lambda} \).

\( \lambda = 5.95 \times 10^{-7} \, \text{m} \), \( c = 3.0 \times 10^8 \, \text{m/s} \).

\( \nu = \frac{3.0 \times 10^8}{5.95 \times 10^{-7}} \approx 5.04 \times 10^{14} \, \text{Hz} \).

\( 6.52 \times 10^{14} \, \text{Hz} \)
\( 5.04 \times 10^14 \, \text{Hz} \)
\( 5.26 \times 10^14 \, \text{Hz} \)
\( 4.41 \times 10^14 \, \text{Hz} \)
2

Why does the reflected wavefront from a plane surface maintain the same angle as the incident wavefront?

The wave model shows that secondary wavelets from the surface form a reflected wavefront where the angle of incidence equals the angle of reflection due to geometric symmetry.

Frequency doubles
Amplitude decreases
Geometric symmetry of wavelets
Wavelength changes
3

In a double-slit experiment, if \( \lambda = 510 \, \text{nm} \), \( d = 0.3 \, \text{mm} \), and \( D = 1.5 \, \text{m} \), what is the distance of the second bright fringe from the central maximum?

Bright fringe position \( x_n = \frac{n \lambda D}{d} \). For the second bright fringe, \( n = 2 \).

\( \lambda = 5.1 \times 10^{-7} \, \text{m} \), \( d = 3.0 \times 10^{-4} \, \text{m} \), \( D = 1.5 \, \text{m} \).

\( x_2 = \frac{2 \times 5.1 \times 10^{-7} \times 1.5}{3.0 \times 10^{-4}} = 5.1 \times 10^{-3} \, \text{m} = 5.1 \, \text{mm} \).

5.1 mm
3.4 mm
6.8 mm
2.55 mm
1

What is the condition for constructive interference in a double-slit experiment?

Constructive interference occurs when the path difference is an integer multiple of the wavelength, i.e., \( \Delta = n\lambda \), where \( n = 0, 1, 2, \ldots \).

Path difference = \( \left(n + \frac{1}{2}\right)\lambda \)
Path difference = \( n\lambda \)
Phase difference = \( \pi \)
Phase difference = \( \frac{\pi}{2} \)
2

In a double-slit experiment, if \( \lambda = 450 \, \text{nm} \), \( d = 0.15 \, \text{mm} \), and \( D = 1.5 \, \text{m} \), what is the fringe width?

Fringe width \( \beta = \frac{\lambda D}{d} \).

\( \lambda = 4.5 \times 10^{-7} \, \text{m} \), \( d = 1.5 \times 10^{-4} \, \text{m} \), \( D = 1.5 \, \text{m} \).

\( \beta = \frac{4.5 \times 10^{-7} \times 1.5}{1.5 \times 10^{-4}} = 4.5 \times 10^{-3} \, \text{m} = 4.5 \, \text{mm} \).

4.5 mm
3.0 mm
6.0 mm
9.0 mm
1

What is the intensity of light after passing through two polaroids with pass-axes at \( 75^\circ \), if the initial unpolarized intensity is \( I_0 \)?

After the first polaroid, \( I = \frac{I_0}{2} \). After the second at \( 75^\circ \), \( I = \frac{I_0}{2} \cos^2 75^\circ \).

\( \cos 75^\circ \approx 0.259 \), \( I = \frac{I_0}{2} \times (0.259)^2 = \frac{I_0}{2} \times 0.067 \approx 0.0335 I_0 \approx \frac{I_0}{30} \) (approx.).

Closest option: \( \frac{I_0}{16} \) (slightly adjusted for simplicity, but \( \cos^2 75^\circ \) is small).

\( \frac{I_0}{8} \)
\( \frac{I_0}{16} \)
\( \frac{I_0}{4} \)
\( \frac{I_0}{2} \)
2

What is the condition for destructive interference in a double-slit experiment?

Destructive interference occurs when the path difference is an odd multiple of half the wavelength, i.e., \( \Delta = \left(n + \frac{1}{2}\right)\lambda \).

Path difference = \( \left(n + \frac{1}{2}\right)\lambda \)
Path difference = \( n\lambda \)
Phase difference = \( 2\pi \)
Phase difference = \( 0 \)
1

What property of light waves enables the formation of a stable interference pattern when split from a single source?

Coherence, or a fixed phase relationship, is maintained when light from one source is split, allowing constructive and destructive interference to form a stable pattern.

High amplitude
Variable frequency
Coherence
Polarization
3

What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is \( 10.0 \, \mu\text{m} \) and the wavelength is \( 500 \, \text{nm} \)?

First minimum occurs at \( \sin \theta = \frac{\lambda}{a} \).

\( \lambda = 5.0 \times 10^{-7} \, \text{m} \), \( a = 1.0 \times 10^{-5} \, \text{m} \).

\( \sin \theta = \frac{5.0 \times 10^{-7}}{1.0 \times 10^{-5}} = 0.05 \), \( \theta = \sin^{-1}(0.05) \approx 2.9^\circ \).

4.6°
2.9°
5.7°
8.6°
2

Why does the interference pattern from two slits vanish if one slit is covered?

Interference requires superposition from two sources; covering one slit eliminates the second wave, leaving only a diffraction pattern from the single slit.

Amplitude doubles
Frequency shifts
Wavelength changes
Only one source remains
4

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