# Q. 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 :

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Solution:

## Path difference = $d \; \sin \theta \approx \; d\theta$ $= 0.1 \times \frac{1}{40} mm = 2500 nm$ or bright fringe, path difference must be integral multiple of $\lambda$ $\therefore \; 2500 = n \lambda_1 = m \lambda_2$ $\therefore \; \lambda_1 = 625 , \lambda_2 = 500$ (from m = 5) (for n = 4)

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