Question

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Derive the Refraction through a rectangular class slab and lateral shift .

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Answer 1

$\sf{\huge{\underline{\red{Answer :-}}}}$

The perpendicular distance between incident & emergent ray when light is incident with parallel phases .

Fig. Refers to the attachment

According to the snells law between medium 1 & medium 2,

➝ $\dfrac{sin\: i1}{sin\: r1}$ = ¹ M 2 -------(1)

According to the Snell's Law between medium2 & medium1 ,

➝ $\dfrac{sin\: r2}{sin\: i2}$ = ² M 1 -------(2)

Multiplying equation (1) & (2) ,

➝ $\dfrac{sin\: i1}{sin\: r1} \times \dfrac{sin\: r2}{sin\: i2}$ = ¹M2 × ²M1

Here, Sin r1 = Sin r2 { Alternate angles are equal }

r1 & r2 the angle of refraction, in fig. make alternate angle

So, we can cancel r1 & r2 .

➝ $\dfrac{sin\: i1}{sin\: i2}$ = 1

➝ sin i1 = sin i2

{ If, sin θ1 = sin θ2

If, sin θ1 = sin θ2 So , θ1 = θ2 }

➝ $\large\bf\pink{\mid{\overline{\underline{i1 = i2 }}}\mid}$

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