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Friday, 9 October

JEE Main 2026 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 5 April 2026, Shift 2, Physics Q42

Question 42 of 75 in this shift, Physics question 17 of 25, Section A.

The maximum intensity in a Young's double slit experiment is I0I_0. Distance between the slits (dd) is 5λ5\lambda, where λ\lambda is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D=10dD=10d is __________.
  1. (1)I04\frac{I_0}{4}
  2. (2)I02\frac{I_0}{2}Official answer
  3. (3)I0I_0
  4. (4)3I04\frac{3I_0}{4}

Official answer

Option 2

NTA final key.