Wave Optics Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the critical angle for light passing from a medium with refractive index 1.9 to air (refractive index 1.0)?

\( \sin i_c = \frac{n_2}{n_1} \), where \( n_1 = 1.9 \), \( n_2 = 1.0 \).

\( \sin i_c = \frac{1.0}{1.9} \approx 0.526 \), \( i_c = \sin^{-1}(0.526) \approx 31.8^\circ \).

36°
31.8°
32.7°
37.3°
2

What is the angular position of the first secondary maximum in a single-slit diffraction pattern if the slit width is \( 9.0 \, \mu\text{m} \) and the wavelength is \( 450 \, \text{nm} \)?

Secondary maxima occur at \( \theta \approx \frac{(n + \frac{1}{2})\lambda}{a} \). For the first secondary maximum, \( n = 1 \).

\( \lambda = 4.5 \times 10^{-7} \, \text{m} \), \( a = 9.0 \times 10^{-6} \, \text{m} \).

\( \sin \theta = \frac{\left(1 + \frac{1}{2}\right) \times 4.5 \times 10^{-7}}{9.0 \times 10^{-6}} = \frac{1.5 \times 4.5 \times 10^{-7}}{9.0 \times 10^{-6}} = 0.075 \), \( \theta = \sin^{-1}(0.075) \approx 4.3^\circ \).

5.7°
4.3°
8.6°
2.9°
2

What allows light to continue propagating as a transverse wave after passing through a polaroid?

The transverse nature of light, with electric fields oscillating perpendicular to propagation, is preserved, but restricted to the polaroid’s pass-axis direction.

Speed increase
Transverse nature
Frequency change
Amplitude reduction
2

What is the intensity at a point in a double-slit experiment where the phase difference is \( 2\pi \), if the maximum intensity is \( 4I_0 \)?

Intensity \( I = 4I_0 \cos^2(\phi/2) \).

For \( \phi = 2\pi \), \( I = 4I_0 \cos^2(\pi) = 4I_0 \times 1 = 4I_0 \).

\( 2I_0 \)
\( I_0 \)
\( 4I_0 \)
\( 0 \)
3

What explains the presence of a central bright fringe in a single-slit diffraction pattern?

All secondary wavelets from the slit interfere constructively at the center (zero angle), producing a bright fringe due to no path difference.

Amplitude decrease
Frequency shift
Wavelength change
Constructive interference
4

What property of light waves explains why the frequency remains constant when light refracts into a denser medium?

Frequency is determined by the source and remains unchanged, while speed and wavelength adjust to the medium’s properties.

Amplitude
Source-dependent frequency
Medium’s density
Wavefront shape
2

What explains the absence of a refracted ray when the angle of incidence exceeds the critical angle?

Beyond the critical angle, the refracted ray would require a sine greater than 1, which is impossible, leading to total internal reflection.

Light’s speed increases
Amplitude decreases
Frequency shifts
Sine of refraction angle exceeds 1
4

What is the intensity at a point in a double-slit experiment where the path difference is \( 3\lambda/4 \), if the maximum intensity is \( 4I_0 \)?

Intensity \( I = 4I_0 \cos^2(\phi/2) \), where \( \phi = \frac{2\pi}{\lambda} \Delta \).

For \( \Delta = \frac{3\lambda}{4} \), \( \phi = \frac{2\pi}{\lambda} \cdot \frac{3\lambda}{4} = \frac{3\pi}{2} \), \( I = 4I_0 \cos^2\left(\frac{3\pi}{4}\right) = 4I_0 \left(\frac{\sqrt{2}}{2}\right)^2 = 4I_0 \times \frac{1}{2} = 2I_0 \).

\( I_0 \)
\( 0 \)
\( 2I_0 \)
\( 4I_0 \)
3

In a double-slit experiment, if \( \lambda = 580 \, \text{nm} \), \( d = 0.2 \, \text{mm} \), and \( D = 1.0 \, \text{m} \), what is the distance of the fourth bright fringe from the central maximum?

Bright fringe position \( x_n = \frac{n \lambda D}{d} \). For the fourth bright fringe, \( n = 4 \).

\( \lambda = 5.8 \times 10^{-7} \, \text{m} \), \( d = 2.0 \times 10^{-4} \, \text{m} \), \( D = 1.0 \, \text{m} \).

\( x_4 = \frac{4 \times 5.8 \times 10^{-7} \times 1.0}{2.0 \times 10^{-4}} = 1.16 \times 10^{-2} \, \text{m} = 11.6 \, \text{mm} \).

11.6 mm
8.7 mm
5.8 mm
14.5 mm
1

In a double-slit experiment, if \( \lambda = 520 \, \text{nm} \), \( d = 0.4 \, \text{mm} \), and \( D = 2.5 \, \text{m} \), what is the fringe width?

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

\( \lambda = 5.2 \times 10^{-7} \, \text{m} \), \( d = 4.0 \times 10^{-4} \, \text{m} \), \( D = 2.5 \, \text{m} \).

\( \beta = \frac{5.2 \times 10^{-7} \times 2.5}{4.0 \times 10^{-4}} = 3.25 \times 10^{-3} \, \text{m} = 3.25 \, \text{mm} \).

3.25 mm
2.6 mm
4.0 mm
1.3 mm
1

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