Q152 / 180 · COMEDK 2024 (Evening Shift) · Physics
In Young's double slit experiment light of wavelength $500 \mathrm{~nm}$ is used to form interference pattern. A uniform glass plate of refractive index 1.5 and thickness $0.1 \mathrm{~mm}$ is introduced in the path of one of the interfering beams. The number of fringes that will shift due to this is
- A400
- Correct answerB100
- C300
- D200
Solution
The shift in the interference pattern in Young's double slit experiment when a glass plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the beams is given by the formula:
$\Delta x = \dfrac{(\mu - 1)tD}{d}$
The number of fringes shifted $N$ is given by the ratio of the shift $\Delta x$ to the fringe width $\beta = \dfrac{\lambda D}{d}$.
$N = \dfrac{\Delta x}{\beta} = \dfrac{(\mu - 1)tD/d}{\lambda D/d} = \dfrac{(\mu - 1)t}{\lambda}$
Given values are $\mu = 1.5$, $t = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} = 10^{-4} \text{ m}$, and $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$.
Substituting these values into the formula:
$N = \dfrac{(1.5 - 1) \times 10^{-4}}{5 \times 10^{-7}}$
$N = \dfrac{0.5 \times 10^{-4}}{5 \times 10^{-7}}$
$N = \dfrac{0.5}{5} \times 10^{3} = 0.1 \times 1000 = 100$
Answer: 100
$\Delta x = \dfrac{(\mu - 1)tD}{d}$
The number of fringes shifted $N$ is given by the ratio of the shift $\Delta x$ to the fringe width $\beta = \dfrac{\lambda D}{d}$.
$N = \dfrac{\Delta x}{\beta} = \dfrac{(\mu - 1)tD/d}{\lambda D/d} = \dfrac{(\mu - 1)t}{\lambda}$
Given values are $\mu = 1.5$, $t = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} = 10^{-4} \text{ m}$, and $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$.
Substituting these values into the formula:
$N = \dfrac{(1.5 - 1) \times 10^{-4}}{5 \times 10^{-7}}$
$N = \dfrac{0.5 \times 10^{-4}}{5 \times 10^{-7}}$
$N = \dfrac{0.5}{5} \times 10^{3} = 0.1 \times 1000 = 100$
Answer: 100
Hence, the correct answer is option: 'B'.