Most Important Selected Qs for JEE AdvancedPhysicsWave Optics
A parallel beam of visible light consisting of wavelengths ₁ and ₂ is incident on a standard YDSE apparatus with d =1 ~mm , D =1 ~m . P is a point on the screen at a distance y from center of screen O . At y=y₁ is the nearest point above O where the two maxima coincide and at y=y₂ is the nearest point above O where the two minima coincide. ₁ & ₂ are fringe width corresponding to wave length ₁ and ₂
Options
- A(A) - (P, R, S), (B) - (P, Q, S), (C) - (P, Q, S)
- B(A) - (P, S), (B) - (Q, S), (C) - (P, Q)
- C(A) - (R, S), (B) - (P, S), (C) - (Q, S)
- D(A) - (P, R, S), (B) - (Q, R, S), (C) - (P, Q, R)
Correct answer
A. (A) - (P, R, S), (B) - (P, Q, S), (C) - (P, Q, S)
Step-by-step solution
aligned & y₁=n₁ ₁=n₂ ₂=L C M of ₁ and ₂ & 2 y₁=2 n₁ ₁=2 n₂ ₂ aligned Hence at this point both maxima again coincide y ₂ ( n ₁- 1 2 ) ₁= ( n ₂- 1 2 ) ₂ ; ₁ ₂ = n ₂- 1 2 n ₁- 1 2 ₁ ₂ = 2 n ₁-1 2 n ₁-1 Which will have a solution. If ₁ ₂ expressed as a proper fraction will be of form odd odd . For (B and C) : ₁ ₂ is of form Odd even . Hence no solution i.e. the two minima will never coincide. For (A) ₁ ₂ is of form odd odd . Hence at some finite y₂ the two minima will coincide. At 2 y₂ the two maxima (and not minima) w