Optics Question 853

Question: Interference fringes were produced using white light in a double slit arrangement. When a mica sheet of uniform thickness of refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves through some a distance. This distance is equal to the width of 30 interference bands if light of wavelength is used. The thickness $ (in\mu m) $ of mica is

Options:

A) 90

B) 12

C) 14

D) 24

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Shift of fringe pattern $ =(\mu -1)\frac{tD}{d} $

$ \therefore \frac{30D(4800\times {{10}^{-10}})}{d}=(0.6)t\frac{D}{d} $

$ 30\times 4800\times {{10}^{-10}}=0.6t $

$ t=\frac{30\times 4800\times {{10}^{-10}}}{0.6}=\frac{1.44\times {{10}^{-5}}}{0.6}=24\times {{10}^{-6}} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक