Question

In youngs double slit experiment a slab of thickness 1.2micrometer and refractive index 1.5 is placed in front of one slit and another slab of thickness t ,refractive index 2.5 is placed infront of another slit if the position of central fringe is unaltered then the thickness t is?

Answer 1

0.72 μm is the thickness of t

Explanation:

Given that,

The thickness of the slab = 1.2μm

Refractive index placed = 1.5

The thickness of another slab = t

Refractive index placed = 2.5

To find,

thickness t without any alteration in the fringe = ?

Procedure:

Because there is no deviation in the placement of the central fringe

Δx equals to zero

so,

The thickness 't' will be determined by

Δx = μ2$t_{2}$ - μ1$t_{1}$

Since Δx = 0. so,

⇒ μ2$t_{2}$ =  μ1$t_{1}$

⇒ 2.5 * $t_{2}$ = 1.2 × 1.5

∵ $t_{2}$ = 0.72 μm

Thus, 0.72 μm is the thickness of t.

Learn more: deviation

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