Q180 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
A monochromatic light of wavelength $6000^{\circ} \mathrm{A}$ is passed through two media A and B of thickness 10 cm and 16 cm respectively. The number of waves in $A$ is $\dfrac{1}{2}$ that of $B$. If the refractive index of $A$ is $\dfrac{4}{3}$, find the refractive index of $B$.
- Correct answerA$\mu_B=\dfrac{5}{3}$
- B$\mu_B=\dfrac{3}{5}$
- C$\mu_B=\dfrac{4}{3}$
- D$\mu_B=\dfrac{3}{2}$
Solution
The number of waves $N$ in a medium of thickness $t$ and refractive index $\mu$ for a wavelength $\lambda$ in vacuum is given by $N = \dfrac{t}{\lambda_{medium}} = \dfrac{t \mu}{\lambda}$.
Given the thickness of medium A is $t_A = 10$ cm and refractive index $\mu_A = \dfrac{4}{3}$, the number of waves in A is $N_A = \dfrac{10 \times (4/3)}{\lambda} = \dfrac{40}{3\lambda}$.
Given the thickness of medium B is $t_B = 16$ cm and refractive index $\mu_B$, the number of waves in B is $N_B = \dfrac{16 \times \mu_B}{\lambda}$.
The problem states that $N_A = \dfrac{1}{2} N_B$, which implies $N_B = 2 N_A$.
Substituting the expressions for $N_A$ and $N_B$:
$\dfrac{16 \mu_B}{\lambda} = 2 \times \dfrac{40}{3\lambda}$
Canceling $\lambda$ from both sides:
$16 \mu_B = \dfrac{80}{3}$
$\mu_B = \dfrac{80}{3 \times 16} = \dfrac{5}{3}$.
Answer: $\mu_B=\dfrac{5}{3}$
Given the thickness of medium A is $t_A = 10$ cm and refractive index $\mu_A = \dfrac{4}{3}$, the number of waves in A is $N_A = \dfrac{10 \times (4/3)}{\lambda} = \dfrac{40}{3\lambda}$.
Given the thickness of medium B is $t_B = 16$ cm and refractive index $\mu_B$, the number of waves in B is $N_B = \dfrac{16 \times \mu_B}{\lambda}$.
The problem states that $N_A = \dfrac{1}{2} N_B$, which implies $N_B = 2 N_A$.
Substituting the expressions for $N_A$ and $N_B$:
$\dfrac{16 \mu_B}{\lambda} = 2 \times \dfrac{40}{3\lambda}$
Canceling $\lambda$ from both sides:
$16 \mu_B = \dfrac{80}{3}$
$\mu_B = \dfrac{80}{3 \times 16} = \dfrac{5}{3}$.
Answer: $\mu_B=\dfrac{5}{3}$
Hence, the correct answer is option: 'A'.