Question:

Young double slit arrangement is placed in a liquid medium of 1.2 refractive index. Distance between the slits and screen is 2.4 m. 
Slit separation is 1 mm. The wavelength of incident light is 5893 Å. The fringe width is:

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Remember to adjust the wavelength for the medium's refractive index when calculating properties like fringe width in optical experiments.
Updated On: Mar 13, 2025
  • 1.17 mm
  • 2.27 mm
  • 1.27 mm
  • 1.5 mm
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The Correct Option is A

Solution and Explanation

Step 1: Convert wavelength to meters and adjust for medium. \[ \lambda = \frac{5893 \times 10^{-10} \, {m}}{1.2} \] Step 2: Apply the formula for fringe width. \[ {Fringe width} \, w = \frac{\lambda \times D}{d} \] Where \( D = 2.4 \, {m} \) and \( d = 1 \times 10^{-3} \, {m} \): \[ w = \frac{4894 \times 10^{-10} \, {m} \times 2.4 \, {m}}{1 \times 10^{-3} \, {m}} \approx 1.17 \, {mm} \]
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