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Standard XII

Physics

YDSE Problems I

A light ray i...

Question

A light ray is incident on an irregular shaped slab of refractive index $\sqrt{2}$ at an angle of $45^{o}$ with the normal on the incline face as shown in the figure. The ray finally emerges from the curved surface in the medium of the refractive index $μ = 1.514$ and passes through point $E$ . If the radius of curved surface is equal to $0.4 m$ , find the distance $O E$ correct upto two decimal places.

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Solution

Given,

Redfractive index $μ_{1} = 1, μ_{2} = \sqrt{2} a n d μ_{3} = 1.514$

Using Snell's law

$μ_{1} sin 45^{\circ} = μ_{2} sin θ$

$θ = 30^{\circ} .$

Therefore, ray moves parallel to axis

$\frac{μ_{3}}{O E} - \frac{μ_{2}}{\infty} = \frac{(μ_{3} - μ_{2})}{R}$

$OE = 6.056 m$

Hence distance OE is $6.06 m$

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Given, Redfractive index μ1=1,μ2=√2andμ3=1.514 Using Snell's law μ1sin45∘=μ2sinθ θ=30∘. Therefore, ray moves parallel to axis μ3OE−μ2∞=(μ3−μ2)R OE=6.056m Hence distance OE is 6.06m