Verified Solution • 2025–26 Session moderate

QNEET-2016-Ph-1051: NEET 2016 Physics - Class 12 NEET-2016-Ph-1051 Solution: The intensity at the maximum in a Young's double slit…

The intensity at the maximum in a Young's double slit experiment is I0I_0. Distance between two slits is d = 5λ5\lambda where λ\lambda is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d? 34I0\frac{3}{4}I_0 I02\frac{I_0}{2} I0I_0 I04\frac{I_0}{4}

Multiple Choice Options

  • A 34I0{3 \over 4}{I_0} 4 3 _ I 0 _
  • B I02{{{I_0}} \over 2} 2 I 0 _ _
  • C I 0
  • D I04{{{I_0}} \over 4} 4 I 0 _ _

Step-by-Step Verified Solution

Path difference S 2 P � S 1 P = ( 50 _ ) 2 + ( 5 _ ) 2 - 50 _ = 0.25 _ S 2 P � S 1 P = _ 4 Phase difference, _ _ = 2 � _ _ _ 4 = � 2 So, resultant intensity at the desired point 'P' is I = I 0 cos 2 _ 2 = I 0 cos 2 � 4 = I 0 2

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