Free Online DPP On Wave Optics - 001

Young's Double Slit Experiment - Problems & Solutions

Q1: In Young's double slit experiment, if the setup is immersed in water (refractive index = 4/3), what will be the new angular fringe width if the initial angular fringe width was 0.2°?

• (A) 0.3°
• (B) 0.15°
• (C) 15°
• (D) 30°

Solution:

The angular fringe width in a medium is given by:

\( \theta' = \frac{\theta}{\mu} \)

Substituting \( \mu = 4/3 \), \( \theta = 0.2° \):

\( \theta' = \frac{0.2}{4/3} = 0.15° \)

Correct Answer: (B) 0.15°

Q2: If Young's double slit experiment is performed in water, keeping the rest of the setup unchanged, what happens to the fringe width?

• (A) Increase in width
• (B) Decrease in width
• (C) Remain unchanged
• (D) Not be formed

Solution:

The fringe width \( \beta \) is given by:

\( \beta = \frac{\lambda D}{d} \)

Since \( \lambda' = \frac{\lambda}{\mu} \) in a medium, fringe width decreases.

Correct Answer: (B) Decrease in width

Q3: A thin mica sheet (thickness \( 2 \times 10^{-6} \) m, refractive index 1.5) is placed in the path of one wave. The wavelength of light used is 5000Å. The shift in central maximum is:

• (A) 2 fringes upward
• (B) 2 fringes downward
• (C) 10 fringes upward
• (D) None of these

Solution:

The shift in fringes is given by:

\( n = \frac{t(\mu - 1)}{\lambda} \)

Substituting values: \( n = \frac{(2 \times 10^{-6})(1.5 - 1)}{5000 \times 10^{-10}} \)

\( n = \frac{10^{-6}}{5000 \times 10^{-10}} = 2 \) fringes.

Correct Answer: (A) 2 fringes upward

Q4: In Young’s Double Slit Experiment, the y-coordinates of central maxima and 10th maxima are 2 cm and 5 cm respectively. When the apparatus is immersed in a liquid of refractive index 1.5, the corresponding y-coordinates will be:

• (A) 2 cm, 7.5 cm
• (B) 3 cm, 6 cm
• (C) 6 cm, 3 cm
• (D) 2 cm, 4 cm

Solution:

Fringe width is given by \( \beta = \frac{\lambda D}{d} \). When immersed in liquid, new fringe width is \( \beta' = \frac{\beta}{\mu} \). Hence, y-coordinates become \( 2 \) cm and \( \frac{5}{1.5} = 4 \) cm.

Correct Answer: (D) 2 cm, 4 cm

Q5: In a Young's double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3 without disturbing the geometrical arrangement, the new fringe width will be:

• (A) 0.30 mm
• (B) 0.40 mm
• (C) 0.53 mm
• (D) 450 micron

Solution:

Fringe width in air is \( \beta = 0.4 \, \text{mm} \). In water, fringe width becomes \( \beta' = \frac{\beta}{\mu} = \frac{0.4}{4/3} = 0.3 \, \text{mm} \).

Correct Answer: (A) 0.30 mm

Q6: Young's double slit experiment is made in a liquid. The 10^th bright fringe lies in liquid where 6^th dark fringe lies in vacuum. The refractive index of the liquid is approximately:

• (A) 1.8
• (B) 1.5
• (C) 1.3
• (D) 1.6

Solution:

For bright fringe in liquid: \( y_{10} = \frac{10 \lambda D}{d \mu} \). For dark fringe in vacuum: \( y_6 = \frac{(6 - 0.5) \lambda D}{d} \). Equating the two, \( \frac{10}{\mu} = 5.5 \), so \( \mu \approx 1.8 \).

Correct Answer: (A) 1.8

Q7: In the Young's double slit experiment, the central maxima is observed to be \( I_0 \). If one of the slits is covered, then intensity at the central maxima will become:

• (A) \( \frac{I_0}{2} \)
• (B) \( \frac{I_0}{\sqrt{2}} \)
• (C) \( \frac{I_0}{4} \)
• (D) \( I_0 \)

Solution:

When both slits are open, intensity at central maxima is \( I_0 \). When one slit is covered, intensity reduces to \( \frac{I_0}{4} \).

Correct Answer: (C) \( \frac{I_0}{4} \)

Q8: In a Young's double slit experiment, the intensity at a point where the path difference is \( \lambda \) (where \( \lambda \) is the wavelength of the light used) is \( I \). If \( I_0 \) denotes the maximum intensity, \( \frac{I}{I_0} \) is equal to:

• (A) \( \frac{1}{\sqrt{2}} \)
• (B) \( \frac{\sqrt{3}}{2} \)
• (C) \( \frac{1}{2} \)
• (D) \( \frac{3}{4} \)

Solution:

Q9: In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double that from the other slit. If \( I_m \) is the maximum intensity, the resultant intensity \( I \) when they interfere at phase difference \( \phi \) is given by:

• (A) \( \frac{I_m}{9} (1 + 8\cos^2(\phi/2)) \)
• (B) \( \frac{I_m}{9} (4 + 5\cos \phi) \)
• (C) \( \frac{I_m}{3} (1 + 2\cos^2(\phi/2)) \)
• (D) \( \frac{I_m}{5} (1 + 4\cos^2(\phi/2)) \)

Solution:

Q10: In Young's double slit experiment, the intensity at a point is \( \frac{1}{4} \) of the maximum intensity. The angular position of this point is:

• (A) \( \arcsin\left(\frac{\lambda}{d}\right) \)
• (B) \( \arcsin\left(\frac{\lambda}{2d}\right) \)
• (C) \( \arcsin\left(\frac{\lambda}{3d}\right) \)
• (D) \( \arcsin\left(\frac{\lambda}{4d}\right) \)

Solution:

Intensity \( I = \frac{I_0}{4} = I_0 \cos^2(\phi/2) \). Solving, \( \cos(\phi/2) = \frac{1}{2} \), so \( \phi/2 = \frac{\pi}{3} \) or \( \phi = \frac{2\pi}{3} \). Path difference \( \Delta x = \frac{\lambda}{2\pi} \cdot \phi = \frac{\lambda}{3} \). Angular position \( \theta = \arcsin\left(\frac{\lambda}{3d}\right) \).

Correct Answer: (C) \( \arcsin\left(\frac{\lambda}{3d}\right) \)

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