Q.49. In a Young's double slit experiment set up, the two slits are kept 0.4. mm apart and screen is

Q: What is the correct answer for JEE Main Physics Q49?

The correct answer is 14. Slit separation $d = 0.4 \text{ mm} = 0.4 \times 10^{-3}$ m.Distance to screen $D = 1$ m.Thickness of sheet $t = 20 \mu\text{m} = 20 \times 10^{-6}$ m.Fringe shift $y_{shift} = 20 \text{ mm} = 20 \times 10^{-3}$ m.The formula for fringe shift due to a thin sheet of refractive index $\mu$ is:$$y_{shift} = \frac{D}{d}(\mu - 1)t$$We are given $y_{shift}$, $D$, $d$, and $t$. We need to find $\mu$.$$20

Q: Q.49. In a Young's double slit experiment set up, the two slits are kept 0.4. mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 m is introduced in front of…

The correct answer is 14. Slit separation $d = 0.4 \text{ mm} = 0.4 \times 10^{-3}$ m.Distance to screen $D = 1$ m.Thickness of sheet $t = 20 \mu\text{m} = 20 \times 10^{-6}$ m.Fringe shift $y_{shift} = 20 \text{ mm} = 20 \times 10^{-3}$ m.The formula for fringe shift due to a thin sheet of refractive index $\mu$ is:$$y_{shift} = \frac{D}{d}(\mu - 1)t$$We are given $y_{shift}$, $D$, $d$, and $t$. We need to find $\mu$.$$20

Question

Q.49. In a Young's double slit experiment set up, the two slits are kept 0.4. mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 m is introduced in front of one of the slits then centre bring fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by  where  is ..........

FirstInTest · Accepted Answer

Slit separation $d = 0.4 	ext{ mm} = 0.4 	imes 10^{-3}$ m.Distance to screen $D = 1$ m.Thickness of sheet $t = 20 \mu	ext{m} = 20 	imes 10^{-6}$ m.Fringe shift $y_{shift} = 20 	ext{ mm} = 20 	imes 10^{-3}$ m.The formula for fringe shift due to a thin sheet of refractive index $\mu$ is:$$y_{shift} = \frac{D}{d}(\mu - 1)t$$We are given $y_{shift}$, $D$, $d$, and $t$. We need to find $\mu$.$$20 	imes 10^{-3} = \frac{1}{0.4 	imes 10^{-3}}(\mu - 1)(20 	imes 10^{-6})$$$$20 	imes 10^{-3} = \frac{20 	imes 10^{-6}}{0.4 	imes 10^{-3}}(\mu - 1)$$$$20 	imes 10^{-3} = \frac{10^{-3}}{0.02}(\mu - 1)$$$$20 	imes 10^{-3} = 50 	imes 10^{-3}(\mu - 1)$$$$20 = 50 (\mu - 1)$$$$\mu - 1 = \frac{20}{50} = 0.4$$$$\mu = 1 + 0.4 = 1.4$$We are given that $\mu = \frac{\alpha}{10}$.So, $1.4 = \frac{\alpha}{10}$.$\alpha = 1.4 	imes 10 = 14$.

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