JEE PYQ: Wave Optics Question 35
Question 35 - 2019 (10 Jan Shift 1)
In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle $\frac{1}{40}$ rad by using light of wavelength $\lambda_1$. When the light of wavelength $\lambda_2$ is used a bright fringe is seen at the same angle in the same set up. Given that $\lambda_1$ and $\lambda_2$ are in visible range (380 nm to 740 nm), their values are:
(1) 625 nm, 500 nm
(2) 380 nm, 525 nm
(3) 380 nm, 500 nm
(4) 400 nm, 500 nm
Show Answer
Answer: (1)
Solution
Path difference $= d\sin\theta = d\theta$
$= 0.1 \times \frac{1}{40}$ mm $= 2500$ nm
For bright fringe, path difference must be integral multiple of $\lambda$.
$\therefore 2500 = n\lambda_1 = m\lambda_2$
$\therefore \lambda_1 = 625$ (for $n = 4$), $\lambda_2 = 500$ (for $m = 5$)
BETA
