Wave Optics Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

In a single-slit diffraction pattern, what happens to the intensity of the central maximum if the slit width is doubled?

Intensity of the central maximum is proportional to \( a^2 \). If \( a \) is doubled, intensity increases by a factor of 4.

Halves
Doubles
Remains the same
Increases four times
4

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

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

Halves
Doubles
Remains the same
Triples
4

What is the intensity of light after passing through two polaroids with pass-axes at \( 60^\circ \), if the intensity after the first is \( I_0 \)?

Using Malus’ law, \( I = I_0 \cos^2 \theta \).

For \( \theta = 60^\circ \), \( \cos 60^\circ = 0.5 \), \( I = I_0 (0.5)^2 = 0.25 I_0 \).

\( \frac{I_0}{2} \)
\( 0.25 I_0 \)
\( I_0 \)
\( 0 \)
2

What is the intensity of light after passing through a polaroid rotated at \( 30^\circ \) relative to the initial polarization direction, if the intensity after the first polaroid is \( I_0 \)?

Using Malus’ law, \( I = I_0 \cos^2 \theta \).

For \( \theta = 30^\circ \), \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), \( I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \times \frac{3}{4} = 0.75 I_0 \).

\( \frac{I_0}{2} \)
\( 0.75 I_0 \)
\( \frac{I_0}{4} \)
\( I_0 \)
2

What is the path difference for the second bright fringe in a double-slit experiment?

Constructive interference occurs at \( \Delta = n\lambda \). For the second bright fringe, \( n = 2 \), so \( \Delta = 2\lambda \).

\( \lambda \)
\( 2\lambda \)
\( \frac{\lambda}{2} \)
\( 3\lambda \)
2

What is the distance of the second dark fringe from the central maximum in a double-slit experiment if \( \lambda = 700 \, \text{nm} \), \( d = 0.35 \, \text{mm} \), and \( D = 1.5 \, \text{m} \)?

For dark fringes, \( x_n = \frac{\left(n + \frac{1}{2}\right) \lambda D}{d} \). Second dark fringe, \( n = 1 \).

\( \lambda = 7.0 \times 10^{-7} \, \text{m} \), \( d = 3.5 \times 10^{-4} \, \text{m} \), \( D = 1.5 \, \text{m} \).

\( x_1 = \frac{\left(1 + \frac{1}{2}\right) \times 7.0 \times 10^{-7} \times 1.5}{3.5 \times 10^{-4}} = \frac{1.5 \times 1.05 \times 10^{-6}}{3.5 \times 10^{-4}} = 4.5 \times 10^{-3} \, \text{m} = 4.5 \, \text{mm} \).

3.0 mm
5.0 mm
4.5 mm
6.0 mm
3

In a double-slit experiment, if \( \lambda = 580 \, \text{nm} \), \( d = 0.25 \, \text{mm} \), and \( D = 2.5 \, \text{m} \), what is the fringe width?

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

\( \lambda = 5.8 \times 10^{-7} \, \text{m} \), \( d = 2.5 \times 10^{-4} \, \text{m} \), \( D = 2.5 \, \text{m} \).

\( \beta = \frac{5.8 \times 10^{-7} \times 2.5}{2.5 \times 10^{-4}} = 5.8 \times 10^{-3} \, \text{m} = 5.8 \, \text{mm} \).

5.8 mm
2.9 mm
7.25 mm
4.64 mm
1

What ensures that light rays from a point source appear to diverge uniformly in all directions?

The isotropic emission of spherical wavefronts from a point source ensures uniform divergence, as wave energy spreads equally in all directions.

Frequency increase
Amplitude variation
Spherical wavefronts
Speed reduction
3

What is the condition for the fourth secondary maximum in a single-slit diffraction pattern?

Secondary maxima occur at \( \theta \approx \frac{(n + \frac{1}{2})\lambda}{a} \). For the fourth secondary maximum, \( n = 4 \), \( \theta \approx \frac{9\lambda}{2a} \).

\( \theta = \frac{7\lambda}{2a} \)
\( \theta = \frac{5\lambda}{a} \)
\( \theta = \frac{3\lambda}{a} \)
\( \theta = \frac{9\lambda}{2a} \)
4

What characteristic of light waves allows a convex lens to transform a plane wave into a converging spherical wave?

The wave nature enables the lens to delay the wavefront variably across its surface, curving it into a spherical shape that converges at the focal point.

Wave nature
Constant speed
High amplitude
Polarization property
1

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