Q147 / 180 · COMEDK 2022 · Physics
In Young's double slit experiment, the two slits are separated by $0.2 \mathrm{~mm}$ and they are $1 \mathrm{~m}$ from the screen. The wavelength of the light used is $500 \mathrm{~nm}$. The distance between 6th maxima and 10 th minima on the screen is closest to
- A$12.0 \mathrm{~mm}$
- B$10.5 \mathrm{~mm}$
- C$14.25 \mathrm{~mm}$
- Correct answerD$8.75 \mathrm{~mm}$
Solution
Given parameters are slit separation $d = 0.2 \mathrm{~mm} = 0.2 \times 10^{-3} \mathrm{~m}$, distance to screen $D = 1 \mathrm{~m}$, and wavelength $\lambda = 500 \mathrm{~nm} = 500 \times 10^{-9} \mathrm{~m}$.
The position of the $n$-th maxima is given by $y_{n, \text{max}} = n \dfrac{D \lambda}{d}$. For the 6th maxima, $n = 6$.
$y_{6, \text{max}} = 6 \times \dfrac{1 \times 500 \times 10^{-9}}{0.2 \times 10^{-3}} = 6 \times 2.5 \times 10^{-3} \mathrm{~m} = 15 \times 10^{-3} \mathrm{~m} = 15 \mathrm{~mm}$.
The position of the $m$-th minima is given by $y_{m, \text{min}} = (m - 0.5) \dfrac{D \lambda}{d}$. For the 10th minima, $m = 10$.
$y_{10, \text{min}} = (10 - 0.5) \times \dfrac{1 \times 500 \times 10^{-9}}{0.2 \times 10^{-3}} = 9.5 \times 2.5 \times 10^{-3} \mathrm{~m} = 23.75 \times 10^{-3} \mathrm{~m} = 23.75 \mathrm{~mm}$.
The distance between the 6th maxima and 10th minima is $|y_{10, \text{min}} - y_{6, \text{max}}| = |23.75 \mathrm{~mm} - 15 \mathrm{~mm}| = 8.75 \mathrm{~mm}$.
Answer: $8.75 \mathrm{~mm}$
The position of the $n$-th maxima is given by $y_{n, \text{max}} = n \dfrac{D \lambda}{d}$. For the 6th maxima, $n = 6$.
$y_{6, \text{max}} = 6 \times \dfrac{1 \times 500 \times 10^{-9}}{0.2 \times 10^{-3}} = 6 \times 2.5 \times 10^{-3} \mathrm{~m} = 15 \times 10^{-3} \mathrm{~m} = 15 \mathrm{~mm}$.
The position of the $m$-th minima is given by $y_{m, \text{min}} = (m - 0.5) \dfrac{D \lambda}{d}$. For the 10th minima, $m = 10$.
$y_{10, \text{min}} = (10 - 0.5) \times \dfrac{1 \times 500 \times 10^{-9}}{0.2 \times 10^{-3}} = 9.5 \times 2.5 \times 10^{-3} \mathrm{~m} = 23.75 \times 10^{-3} \mathrm{~m} = 23.75 \mathrm{~mm}$.
The distance between the 6th maxima and 10th minima is $|y_{10, \text{min}} - y_{6, \text{max}}| = |23.75 \mathrm{~mm} - 15 \mathrm{~mm}| = 8.75 \mathrm{~mm}$.
Answer: $8.75 \mathrm{~mm}$
Hence, the correct answer is option: 'D'.