Practice free →
HomeJEE MainPhysicsWave Optics › In Young's double-slit experiment, fringe width …

In Young's double-slit experiment, fringe width $\beta = \dfrac{\lambda D}{d}$. If the screen distance $D$ is doubled and the slit separation $d$ is halved, the new fringe width is:

A$\beta$
B$2\beta$
C$4\beta$
D$\dfrac{\beta}{4}$
Answer & Solution
Correct answer: C. $4\beta$
Plug in directly. New $\beta' = \dfrac{\lambda \cdot (2D)}{(d/2)} = \dfrac{2 \lambda D}{d/2} = 4 \cdot \dfrac{\lambda D}{d} = 4\beta$. Intuition: increasing $D$ stretches the geometry linearly, and decreasing $d$ shrinks the angular separation linearly. Both effects widen the fringes, and together they multiply.
Solve this in the app — JEE Main practice & 24k+ MCQs →
Related questions