Problem

A plate of thickness

t

made of a material of refractive index

\mu

is placed in front of one of the slits in a double slit experiment. a. Find the change in the optical path due to introduction of the plate. b. What should be the minimum thickness

t

which will make the intensity at the centre of the fringe pattern zero? Wavelength of the light used is

\lambda

, neglect any absorption of light in the plate. (a)

(\mu - 1)t, \frac{\lambda}{2(\mu - 1)}

(b)

2(\mu - 1)t, \frac{\lambda}{\mu - 1}

(c)

(\mu - 1)t, \frac{\lambda}{\mu - 1}

(d)

(\mu - 1) \frac{t}{2}, \frac{\lambda}{\mu - 1}

Intuitive explanation

The core concept here is optical path length — how a material's refractive index effectively "slows down" light and changes the phase relationship between the two slits. When you insert a plate of refractive index $\mu$ in front of one slit, you're not just blocking light; you're introducing an extra optical path that shifts one wave relative to the other. The key insight is that the optical path through the plate is $\mu t$ (not just $t$ ), so the additional phase delay compared to traveling the same distance through air is $(\mu - 1)t$ . This shifts the entire interference pattern, and you can find the new fringe positions by accounting for this extra path difference.

Intuitive explanation

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