Refraction through a rectangular glass slab - Physics - eSaral

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Refraction through slab:

The Refractive index & thickness of the glass slab is µ & t respectively. One light ray AB incidents on a slab, Displacement produced, in emergent ray due to refraction. $x=\frac{t \sin (i-r)}{\cos r}=t \sec r \sin (i-r)$

(a) When an object is in denser medium & observer in rarer medium:

The thickness of a denser medium is t, in which an object is at a point O. Due to refraction, the image may be seen at a point I. Refractive index

$\mu=\frac{\text { Real depth }}{\text { Virtualdepth }}=\frac{\mathrm{AO}}{\mathrm{Al}}=\frac{\mathrm{t}}{\mathrm{Al}}$

Virtual depth $=\frac{t}{\mu}$

Virtual displacement $(\mathrm{OI})=\mathrm{OA}-\mathrm{Al}=\mathrm{t}\left(1-\frac{1}{\mathrm{\mu}}\right)$

(b) Refraction through a successive slab of different thickness & refractive index.

Virtual depth $(\mathrm{Al})=\frac{\mathrm{t}_{1}}{\mu_{1}}+\frac{\mathrm{t}_{2}}{\mu_{2}}+\frac{\mathrm{t}_{3}}{\mu_{3}}+\ldots$

Virtual displacement (Ol) $=\mathrm{t}_{1}\left(1-\frac{1}{\mu_{1}}\right)+\mathrm{t}_{2}\left(1-\frac{1}{\mu_{2}}\right)+\mathrm{t}_{3}\left(1-\frac{1}{\mu_{3}}\right)+$............

(c) When object & observer both are in rarer medium.

Let observer is in air & object is at a point O in air, as shown in the figure. A glass is there in between observer & object. Images forms at point I Refractive index of glass is $\mu$. Virtual displacement $=\mathrm{Ol}=\left(\mathrm{t}-\frac{1}{\mathrm{\mu}}\right)$

(d) When an object in a rarer medium & Observer in the denser medium.

The Refractive index of water is $\mu .$ Observer is in water, Image may be seen at a point I when an object at a point O is viewed.

$\frac{\text { Re al height }}{\text { Virtual height }}=\frac{1}{\mu}$

Virtual displacement $(\mathrm{O} \mid)=\mathrm{Al}-\mathrm{AO}=(\mu-1) \mathrm{AO}$

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