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(D) The number of waves $N$ in a medium of thickness $t$ and refractive index $\mu$ is given by $N = \frac{t}{\lambda_m} = \frac{t \cdot \mu}{\lambda}

Vedclass · Accepted Answer

$1.7$

Vedclass · Answer

(D) The number of waves $N$ in a medium of thickness $t$ and refractive index $\mu$ is given by $N = \frac{t}{\lambda_m} = \frac{t \cdot \mu}{\lambda}$,where $\lambda$ is the wavelength in vacuum.For medium $A$,thickness $t_A = 3x$ and $\mu_A = 1.5$. So,$N_A = \frac{3x \cdot 1.5}{\lambda} = \frac{4.5x}{\lambda}$.For the combination of $B$ and $C$,thickness $t_B = x$ and $t_C = 2x$. Let $\mu_B = \mu$. The number of waves is $N_{B+C} = N_B + N_C = \frac{x \cdot \mu}{\lambda} + \frac{2x \cdot 1.4}{\lambda} = \frac{x(\mu + 2.8)}{\lambda}$.Given $N_A = N_{B+C}$,we have:$\frac{4.5x}{\lambda} = \frac{x(\mu + 2.8)}{\lambda}$$4.5 = \mu + 2.8$$\mu = 4.5 - 2.8 = 1.7$.Thus,the refractive index of $B$ is $1.7$.

Consider slabs of three media $A, B$ and $C$ arranged as shown in the figure. The refractive index $(R.I.)$ of $A$ is $1.5$ and that of $C$ is $1.4$. If the number of waves in $A$ is equal to the number of waves in the combination of $B$ and $C$,then the refractive index of $B$ is:

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