Wave Optics Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In a single-slit diffraction pattern, what happens to the central maximum’s width if the wavelength is quadrupled?

Angular width \( 2\theta = \frac{2\lambda}{a} \). If \( \lambda \) is quadrupled, \( 2\theta \) increases four times.

Halves
Doubles
Triples
Increases four times
4

In a double-slit experiment, if the slit separation is doubled, what happens to the fringe width?

Fringe width \( \beta = \frac{\lambda D}{d} \). If \( d \) is doubled, \( \beta \) is halved.

Doubles
Triples
Remains the same
Halves
4

Why does the wave theory require light to propagate as transverse waves to explain polarization?

Only transverse waves have oscillations perpendicular to propagation that can be filtered in specific directions, unlike longitudinal waves, enabling polarization.

Speed variation
Frequency stability
Amplitude consistency
Perpendicular oscillations
4

What is the refractive index of a medium if the critical angle for light passing into air is \( 36.9^\circ \)?

\( \sin i_c = \frac{n_2}{n_1} \), where \( n_2 = 1.0 \) (air), \( i_c = 36.9^\circ \).

\( \sin 36.9^\circ \approx 0.6 \), \( n_1 = \frac{1.0}{0.6} \approx 1.67 \).

1.5
1.33
1.67
1.8
3

What is the shape of the wavefront emitted by a point source in a uniform medium close to the source?

A point source in a uniform medium emits spherical wavefronts, especially close to the source.

Spherical
Plane
Cylindrical
Tilted
1

What is the resultant amplitude of two coherent waves of amplitude \( a \) with a phase difference of \( 6\pi \)?

Resultant amplitude \( A = 2a \cos(\phi/2) \).

For \( \phi = 6\pi \), \( A = 2a \cos(3\pi) = 2a \times (-1) = -2a \), magnitude \( 2a \).

\( a \)
\( a\sqrt{2} \)
\( 0 \)
\( 2a \)
4

What is the shape of the wavefront after a plane wave passes through a thin prism?

A plane wave passing through a thin prism gets delayed variably, resulting in a tilted plane wavefront.

Tilted plane
Spherical
Cylindrical
Plane
1

What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width is \( 3.0 \, \mu\text{m} \) and the wavelength is \( 450 \, \text{nm} \)?

Angular width \( 2\theta = \frac{2\lambda}{a} \).

\( \lambda = 4.5 \times 10^{-7} \, \text{m} \), \( a = 3.0 \times 10^{-6} \, \text{m} \).

\( \sin \theta = \frac{\lambda}{a} = \frac{4.5 \times 10^{-7}}{3.0 \times 10^{-6}} = 0.15 \), \( \theta = \sin^{-1}(0.15) \approx 8.6^\circ \), \( 2\theta \approx 17.2^\circ \).

15°
17.2°
20°
10°
2

In a double-slit experiment, if the intensity at the central maximum is \( 4I_0 \), what is the intensity where the path difference is \( \lambda/4 \)?

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

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

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

Why does the intensity of light transmitted through two polaroids drop to zero when their pass-axes are perpendicular?

When pass-axes are perpendicular, the electric field component along the second polaroid’s axis is zero (cos 90° = 0), blocking all light per Malus’ law.

Frequency cancels
Wavelength doubles
Amplitude increases
No electric field component passes
4

What is the angular position of the third minimum in a single-slit diffraction pattern if the slit width is \( 6.0 \, \mu\text{m} \) and the wavelength is \( 480 \, \text{nm} \)?

Minima occur at \( \sin \theta = \frac{n\lambda}{a} \). For the third minimum, \( n = 3 \).

\( \lambda = 4.8 \times 10^{-7} \, \text{m} \), \( a = 6.0 \times 10^{-6} \, \text{m} \).

\( \sin \theta = \frac{3 \times 4.8 \times 10^{-7}}{6.0 \times 10^{-6}} = 0.24 \), \( \theta = \sin^{-1}(0.24) \approx 13.9^\circ \).

17.5°
13.9°
20°
10°
2

Why does the intensity of transmitted light through a polaroid decrease when rotated, even if the incident light is unpolarized?

Unpolarized light becomes polarized after the first polaroid, and the second polaroid’s pass-axis alignment determines the transmitted component, reducing intensity as the angle increases.

Frequency reduction
Polarization of unpolarized light
Amplitude doubling
Speed variation
2

What happens to the intensity of light when it passes through a polaroid and the polaroid is rotated by \( 90^\circ \) from its initial position?

For unpolarized light, intensity after one polaroid is \( \frac{I_0}{2} \). Rotating by \( 90^\circ \) from the pass-axis gives \( I = \frac{I_0}{2} \cos^2 90^\circ = 0 \).

It reduces to zero from its initial value.

Increases
Remains the same
Doubles
Reduces to zero
4

Why does the intensity of light remain unchanged in terms of energy when it undergoes interference or diffraction?

Interference and diffraction redistribute light energy without loss, as bright and dark regions balance out, conserving total energy.

Energy is redistributed, not lost
Amplitude increases overall
Frequency changes
Wavelength decreases
1

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