Q142 / 160 · TS EAMCET 2023 (12 May Shift 1) · Physics
When Young's double slit experiment is performed in air, the Y-coordinates of central maxima and $10^{\text {th }}$ maxima are $2 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. If the apparatus is immersed in a liquid of refractive index 1.5 , the corresponding Y-coordinates will be
- A$2 \mathrm{~cm}, 7.5 \mathrm{~cm}$
- B$3 \mathrm{~cm}, 6 \mathrm{~cm}$
- Correct answerC$2 \mathrm{~cm}, 4 \mathrm{~cm}$
- D$\frac{4}{3} \mathrm{~cm}, \frac{10}{3} \mathrm{~cm}$
Solution
For YDSE apparatus in air
For central maxima
$\begin{aligned}
& \frac{y_0 d}{D}-d \sin \theta=0 \\
& \Rightarrow \sin \theta=\frac{y_0}{D}=\frac{2}{D} ... (i)
\end{aligned}$
For $10^{\text {th }}$ maxima
$\begin{aligned}
& \frac{\mathrm{y}_{10} \mathrm{~d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda \\
& \Rightarrow \frac{5 \mathrm{~d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda
\end{aligned}$
From equation (i), we have
$\begin{aligned}
& \frac{5 d}{D}-\frac{2 d}{D}=10 \lambda \\
& \Rightarrow 10 \lambda=\frac{3 d}{D} \Rightarrow \lambda=\frac{3 d}{10 D}
\end{aligned}$
Now, when the YDSE apparatus is immersed in a liquid of refractive index 1.5 , then it's wave length $\lambda^{\prime}=\frac{\lambda}{1.5}$
As the y coordinate of central maximum doesn't depend upon wavelength, therefore it will remain same as in air, $\mathrm{y}_0^{\prime}=\mathrm{y}_0=2 \mathrm{~cm}$
For $10^{\text {th }}$ maxina
$\begin{aligned}
& \frac{\mathrm{y}_{10}^{\prime} \mathrm{d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda^{\prime} \\
& \Rightarrow \frac{\mathrm{y}_{10}^{\prime} \mathrm{d}}{\mathrm{D}}-\frac{\mathrm{d} \times 2}{\mathrm{D}}=\frac{10 \times 3 \mathrm{~d}}{1.510 \mathrm{D}} \\
& \Rightarrow \mathrm{y}_{10}^{\prime}=4 \mathrm{~cm}
\end{aligned}$
Hence, the correct answer is option: 'C'.