Q.50. In a double slit experiment, when one of the slits is covered by a transparent mica sheet of r

Q: What is the correct answer for JEE Main Physics Q50?

The correct answer is 5625. Sol. The shift in the central maximum due to the introduction of a thin sheet of thickness t and refractive index $\mu$ is given by $\Delta x = \frac{D}{d} (\mu - 1)t$. This shift is equal to the position of the $7^{th}$ bright fringe, which is $y_7 = 7\frac{\lambda D}{d}$. So, $\frac{D}{d} (\mu - 1)t = 7\frac{\lambda D}{d}$. $(\mu - 1)t = 7\lambda$. Given $\mu = 1.56$, $\lambda = 450 \text{ nm} =

Q: Q.50. In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of bright fringe, obtained with…

The correct answer is 5625. Sol. The shift in the central maximum due to the introduction of a thin sheet of thickness t and refractive index $\mu$ is given by $\Delta x = \frac{D}{d} (\mu - 1)t$. This shift is equal to the position of the $7^{th}$ bright fringe, which is $y_7 = 7\frac{\lambda D}{d}$. So, $\frac{D}{d} (\mu - 1)t = 7\frac{\lambda D}{d}$. $(\mu - 1)t = 7\lambda$. Given $\mu = 1.56$, $\lambda = 450 \text{ nm} =

Question

Q.50. In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index 1.56, the central fringe shifts to the position of  bright fringe, obtained with both slits uncovered. If the light source wavelength is 450 nm, the thickness of mica sheet is . The value of  is __________

FirstInTest · Accepted Answer

Sol. The shift in the central maximum due to the introduction of a thin sheet of thickness t and refractive index $\mu$ is given by $\Delta x = \frac{D}{d} (\mu - 1)t$. This shift is equal to the position of the $7^{th}$ bright fringe, which is $y_7 = 7\frac{\lambda D}{d}$. So, $\frac{D}{d} (\mu - 1)t = 7\frac{\lambda D}{d}$. $(\mu - 1)t = 7\lambda$. Given $\mu = 1.56$, $\lambda = 450 	ext{ nm} = 450 	imes 10^{-9} 	ext{ m}$. $(1.56 - 1)t = 7 	imes 450 	imes 10^{-9}$. $0.56t = 3150 	imes 10^{-9}$. $t = \frac{3150 	imes 10^{-9}}{0.56} = \frac{3150}{0.56} 	imes 10^{-9} = 5625 	imes 10^{-9} 	ext{ m}$. We are given the thickness is $a 	imes 10^{-6} 	ext{ m}$. So, $a 	imes 10^{-6} = 5625 	imes 10^{-9}$. $a = 5625 	imes 10^{-3} = 5.625$. The solution states $t = 5625 	ext{ nm}$. If $t = 5625 	ext{ nm} = 5625 	imes 10^{-9} 	ext{ m}$. Then $a 	imes 10^{-6} = 5625 	imes 10^{-9}$. $a = 5625 	imes 10^{-3} = 5.625$. However, the answer is given as 5625. This implies the question meant $a$ in nanometers, or the unit in the question is $a 	imes 10^{-9} 	ext{ m}$. If the thickness is $a 	imes 10^{-9} 	ext{ m}$, then $a = 5625$. Given the answer is 5625, it's likely the question meant $a 	imes 10^{-9} 	ext{ m}$ or $a$ is in nm. Assuming the question meant $a$ is the value in nm, then $a = 5625$. Or if the question meant $a 	imes 10^{-9}$ m, then $a=5625$. If it's $a 	imes 10^{-6}$ m, then $a=5.625$. Given the integer answer 5625, I will assume the question implies $a$ is the value in nanometers, or there's a typo in the power of 10. I will use $a=5625$.

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