Two transparent slabs have the same thickness | vedclass

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(C) Let $\lambda$ be the wavelength of the monochromatic light in air. The wavelength of light in a medium with refractive index $\mu$ is given by $\l

Vedclass · Accepted Answer

$1.3$

Vedclass · Answer

(C) Let $\lambda$ be the wavelength of the monochromatic light in air. The wavelength of light in a medium with refractive index $\mu$ is given by $\lambda_m = \frac{\lambda}{\mu}$.The number of wavelengths $N$ in a slab of thickness $d$ is given by $N = \frac{d}{\lambda_m} = \frac{d \cdot \mu}{\lambda}$.For slab $X$ (thickness $t$,refractive index $1.5$):$N_X = \frac{t \cdot 1.5}{\lambda}$.For the second slab,it consists of two parts $Y$ and $Z$ with thicknesses $t_Y = \frac{t}{3}$ and $t_Z = \frac{2t}{3}$. Let the refractive index of $Y$ be $\mu_Y$.$N_{YZ} = \frac{t_Y \cdot \mu_Y}{\lambda} + \frac{t_Z \cdot 1.6}{\lambda} = \frac{(t/3) \cdot \mu_Y}{\lambda} + \frac{(2t/3) \cdot 1.6}{\lambda}$.Given that the number of wavelengths is the same in both slabs $(N_X = N_{YZ})$:$\frac{1.5 t}{\lambda} = \frac{t \cdot \mu_Y}{3 \lambda} + \frac{2t \cdot 1.6}{3 \lambda}$.Dividing both sides by $t/\lambda$:$1.5 = \frac{\mu_Y}{3} + \frac{3.2}{3}$.Multiplying by $3$:$4.5 = \mu_Y + 3.2$.$\mu_Y = 4.5 - 3.2 = 1.3$.

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