Q152 / 180 · COMEDK 2026 (9 May Shift 1) · Physics
In a Young's double slit experiment, the slits are separated by 0.5 mm. Fringes are obtained on a screen which is placed at distance 1 m away from the slits. When the screen is moved 7 cm farther away, the fringe width changes by 63 $\mu$m. The wavelength of light used in the experiment will be:
- Correct answerA$4.5 \times 10^{-7}\ m$
- B$0.5 \times 10^{-9}\ m$
- C$4.5 \times 10^{-6}\ m$
- D$0.5 \times 10^{-7}\ m$
Solution
The fringe width in a Young's double slit experiment is given by $\beta = \dfrac{\lambda D}{d}$
When the screen is moved by a distance $\Delta D$, the change in fringe width is $\Delta \beta = \dfrac{\lambda \Delta D}{d}$
Rearranging the formula to solve for the wavelength $\lambda$:
$\lambda = \dfrac{\Delta \beta \cdot d}{\Delta D}$
Substituting the given values $\Delta \beta = 63 \times 10^{-6}\ m$, $d = 0.5 \times 10^{-3}\ m$, and $\Delta D = 7 \times 10^{-2}\ m$:
$\lambda = \dfrac{63 \times 10^{-6} \times 0.5 \times 10^{-3}}{7 \times 10^{-2}}$
$\lambda = \dfrac{31.5 \times 10^{-9}}{7 \times 10^{-2}}$
$\lambda = 4.5 \times 10^{-7}\ m$
Answer: $4.5 \times 10^{-7}\ m$
When the screen is moved by a distance $\Delta D$, the change in fringe width is $\Delta \beta = \dfrac{\lambda \Delta D}{d}$
Rearranging the formula to solve for the wavelength $\lambda$:
$\lambda = \dfrac{\Delta \beta \cdot d}{\Delta D}$
Substituting the given values $\Delta \beta = 63 \times 10^{-6}\ m$, $d = 0.5 \times 10^{-3}\ m$, and $\Delta D = 7 \times 10^{-2}\ m$:
$\lambda = \dfrac{63 \times 10^{-6} \times 0.5 \times 10^{-3}}{7 \times 10^{-2}}$
$\lambda = \dfrac{31.5 \times 10^{-9}}{7 \times 10^{-2}}$
$\lambda = 4.5 \times 10^{-7}\ m$
Answer: $4.5 \times 10^{-7}\ m$
Hence, the correct answer is option: 'A'.