Wave Optics Chapter-Wise Test 5

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What is the maximum intensity of light transmitted through three polaroids when the first and third are crossed, and the middle one is rotated, if the initial intensity is \( I_0 \)?

For crossed polaroids, intensity after the second polaroid is \( I_0 \cos^2 \theta \), and after the third (at \( 90^\circ - \theta \)) is \( I = I_0 \cos^2 \theta \sin^2 \theta = \frac{I_0}{4} \sin^2 2\theta \).

Maximum occurs at \( \theta = 45^\circ \), so \( I = \frac{I_0}{4} \).

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

What is the intensity of unpolarised light after passing through a single polaroid, if the initial intensity is \( I_0 \)?

Unpolarised light has its intensity reduced to half after passing through a polaroid, so \( I = \frac{I_0}{2} \).

\( \frac{I_0}{2} \)
\( I_0 \)
\( \frac{I_0}{4} \)
\( 0 \)
1

What is the nature of unpolarized light?

Unpolarized light has electric vectors oscillating in all directions in the plane perpendicular to the propagation direction, randomly changing over time.

Oscillates in one direction
Longitudinal
Stationary
Randomly oscillating in all transverse directions
4

What happens to the refracted ray when the angle of incidence exceeds the critical angle?

When the angle of incidence exceeds the critical angle, no refraction occurs, and the ray undergoes total internal reflection.

Bends towards the normal
Bends away from the normal
Undergoes total internal reflection
Passes undeflected
3

What is the angular position of the second minimum in a single-slit diffraction pattern if the slit width is \( 8.0 \, \mu\text{m} \) and the wavelength is \( 640 \, \text{nm} \)?

Minima occur at \( \sin \theta = \frac{n\lambda}{a} \). For the second minimum, \( n = 2 \).

\( \lambda = 6.4 \times 10^{-7} \, \text{m} \), \( a = 8.0 \times 10^{-6} \, \text{m} \).

\( \sin \theta = \frac{2 \times 6.4 \times 10^{-7}}{8.0 \times 10^{-6}} = 0.16 \), \( \theta = \sin^{-1}(0.16) \approx 9.2^\circ \).

13.9°
9.2°
4.6°
23.6°
2

What is the phase difference corresponding to a path difference of \( 3\lambda/4 \) in a double-slit experiment?

Phase difference \( \phi = \frac{2\pi}{\lambda} \Delta \).

For \( \Delta = \frac{3\lambda}{4} \), \( \phi = \frac{2\pi}{\lambda} \cdot \frac{3\lambda}{4} = \frac{3\pi}{2} \).

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

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

Secondary maxima occur at \( \theta \approx \frac{(n + \frac{1}{2})\lambda}{a} \). For the second secondary maximum, \( n = 2 \), \( \theta \approx \frac{5\lambda}{2a} \).

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

What is the condition for total internal reflection to occur?

Total internal reflection occurs when the angle of incidence exceeds the critical angle, and light travels from a denser to a rarer medium.

Angle of incidence < Critical angle
Angle of incidence = 90°
Light travels from rarer to denser medium
Angle of incidence > Critical angle
4

Why does the diffraction pattern of a single slit show a central maximum broader than its secondary maxima?

The central maximum results from constructive interference of all secondary wavelets in phase, while secondary maxima involve partial cancellations, reducing their width and intensity.

Frequency variation
Amplitude increase
Wavelength change
Constructive interference of wavelets
4

What is the wavelength of light in a medium with refractive index 1.65 if its wavelength in air is \( 660 \, \text{nm} \)?

Wavelength in a medium \( \lambda_m = \frac{\lambda_{\text{air}}}{n} \).

Given \( \lambda_{\text{air}} = 660 \, \text{nm} \), \( n = 1.65 \), \( \lambda_m = \frac{660}{1.65} = 400 \, \text{nm} \).

400 nm
660 nm
500 nm
330 nm
1

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