Light falls at normal incidence on one face of a glass prism of refractive index -2- Then the angle of emergence when the angle of the prism is 45 -

The correct option is D 90 ^∘ μ× sin 45=sin r× 1 √(2)×1√(2)=sin r ⇒ r=90^∘ ...

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Standard VII

Physics

Reflection

Light falls a...

Question

Light falls at normal incidence on one face of a glass prism of refractive index $\sqrt{2}$ . Then the angle of emergence when the angle of the prism is 45 $^{\circ}$

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Solution

The correct option is D 90 $^{\circ}$
$μ \times s i n 45 = s i n r \times 1$
$\sqrt{2} \times \frac{1}{\sqrt{2}} = s i n r$

$\Rightarrow r = 90^{\circ}$

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Light falls at normal incidence on one face of a glass prism of refractive index √2. Then the angle of emergence when the angle of the prism is 45∘

The correct option is D 90∘μ×sin45=sinr×1√2×1√2=sinr⇒r=90∘

Light falls at normal incidence on one face of a glass prism of refractive index $\sqrt{2}$ . Then the angle of emergence when the angle of the prism is 45 $^{\circ}$

The correct option is D 90 $^{\circ}$
$μ \times s i n 45 = s i n r \times 1$
$\sqrt{2} \times \frac{1}{\sqrt{2}} = s i n r$

$\Rightarrow r = 90^{\circ}$