In Young's double slit experiment, the slits are of equal width and maximum intensity of light on the screen is found to be Im. Now one of the slits is covered with a thin sheet of thickness $$15$$λ$$4$$, made of material of refractive index $$1.2$$. Then intensity of light at the mid point of the source is

Question

In Young's double slit experiment, the slits are of equal width and maximum intensity of light on the screen is found to be Im. Now one of the slits is covered with a thin sheet of thickness $$15$$λ$$4$$, made of material of refractive index $$1.2$$. Then intensity of light at the mid point of the source is

Infinity Learn · Accepted Answer

If $$I0$$ be the intensity of light energy from each slit, then $$Im=4I0$$.When one of the slits is covered, path difference between rays at the mid point of the screen $$=$$μ−$$1t=1.2$$−$$1$$×$$15$$λ$$4=3$$λ$$4$$∴ ϕ$$=phase$$ difference $$=2$$πλ×$$3$$λ$$4=3$$$$π$$$$2$$∴ At the mid point, $$I=Im$$.$$cos2$$ϕ$$2=Imcos23$$$$π$$$$4=ImIV22=Im2$$

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In Young's double slit experiment, the slits are of equal width and maximum intensity of light on the screen is found to be Im. Now one of the slits is covered with a thin sheet of thickness 15λ4, made of material of refractive index 1.2. Then intensity of light at the mid point of the source is

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