JEE PYQ: Wave Optics Question 23
Question 23 - 2020 (09 Jan Shift 2)
In a Young’s double slit experiment 15 fringes are observed on a small portion of the screen when light of wavelength 500 nm is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in nm) __________.
Show Answer
Answer: 750
Solution
Fringe width, $\beta = \frac{\lambda D}{d}$ where, $\lambda$ = wavelength, $D$ = distance of screen from slits, $d$ = distance between slits.
ATQ: $15 \times \frac{\lambda_1 D}{d} = 10 \times \frac{\lambda_2 D}{d}$
$\Rightarrow 15\lambda_1 = 10\lambda_2$
$\Rightarrow \lambda_2 = 1.5\lambda_1 = 1.5 \times 500$ nm
$\Rightarrow \lambda_2 = 750$ nm
BETA
