In YDSE, films of thickness t₁ and t₂ are in front of slits S₁ and S₂ in experiment 1; a first minimum occurs at O. In experiment 2 both films are placed in front of S₁ and the second maximum occurs at O. Same refractive index. If t₂/t₁ = x/10, calculate x. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

In YDSE, films of thickness t₁ and t₂ are in front of slits S₁ and S₂ in experiment 1; a first minimum occurs at O. In experiment 2 both films are placed in front of S₁ and the second maximum occurs at O. Same refractive index. If t₂/t₁ = x/10, calculate x. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — On central screen line, inserting film t gives optical delay (μ−1)t. For the first setup, path difference magnitude is (μ−1)(t₁−t₂) = λ/2. Step 2 — For second setup, both films are in front of same slit; second bright fringe requires (μ−1)(t₁+t₂) = 2λ. Step 3 — Divide: (t₁+t₂)/(t₁−t₂) = 4, leading to t₁+t₂ = 4t₁−4t₂. Step 4 — Thus 5t₂ = 3t₁, giving t₂/t₁ = 3/5 = 0.6. Step 5 — t₂/t₁ = x/10, so x = 6.