Question:
A thin glass plate of thickness \(\frac{2500}{3}\lambda\) (\(\lambda\) is wavelength of light used) and refractive index \(\mu = 1.5\) is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :
A thin glass plate of thickness \(\frac{2500}{3}\lambda\) (\(\lambda\) is wavelength of light used) and refractive index \(\mu = 1.5\) is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :
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The plate adds an extra optical path \((\mu-1)t\); convert to a phase difference, then use \(I = I_{max}\cos^2(\phi/2)\).
Updated On: Nov 4, 2026
- 2 : 1
- 1 : 4
- 4 : 1
- 4 : 3
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The Correct Option is C
Solution and Explanation
Extra path \(= (\mu - 1)t = 0.5\times\dfrac{2500}{3}\lambda = \dfrac{1250}{3}\lambda = 416\lambda + \dfrac{2}{3}\lambda\).
Effective phase difference \(\phi = 2\pi\times\dfrac{2}{3} = \dfrac{4\pi}{3}\) (whole waves do not matter).
Before insertion: central maximum, \(I_1 = I_{max} = 4I_0\). After insertion: \(I_2 = 4I_0\cos^2\dfrac{\phi}{2} = 4I_0\cos^2\dfrac{2\pi}{3} = 4I_0\times\dfrac14 = I_0\).
\(I_1 : I_2 = 4I_0 : I_0 = 4 : 1\).
Answer: (3) 4 : 1
Effective phase difference \(\phi = 2\pi\times\dfrac{2}{3} = \dfrac{4\pi}{3}\) (whole waves do not matter).
Before insertion: central maximum, \(I_1 = I_{max} = 4I_0\). After insertion: \(I_2 = 4I_0\cos^2\dfrac{\phi}{2} = 4I_0\cos^2\dfrac{2\pi}{3} = 4I_0\times\dfrac14 = I_0\).
\(I_1 : I_2 = 4I_0 : I_0 = 4 : 1\).
Answer: (3) 4 : 1
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