The angle of a prism is 60-circ- If the refractive index of the material of the prism is sqrt-2- the angle of minimum deviation is -45-circ-60-circ-30-circ-sin-1-dfrac-2-3-

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Toppr Admin · Accepted Answer

-x3BC-sin-D-A2-sin-A-2-xA0- -xA0-x3BC-2sin-D-602-xA0- -xA0-2sin-D-602-xA0- -xA0-1-x221A-2-sinD-60-x2218-2-xA0- -xA0-45-x2218-D-60-x2218-2-xA0- -xA0-D-90-x2212-60-x2218-xA0- -xA0-30-x2218-

The angle of a prism is 60 $^{\circ}$ . If the refractive index of the material of the prism is $\sqrt{2}$ the angle of minimum deviation is :
45 $^{\circ}$
60 $^{\circ}$
30 $^{\circ}$
sin $^{- 1} (\frac{2}{3})$

$μ = \frac{s i n (\frac{D + A}{2})}{s i n (A / 2)}$

$μ = 2 s i n (\frac{D + 60}{2})$

$= 2 s i n (\frac{D + 60}{2})$

$\frac{1}{\sqrt{2}} = s i n \frac{D + 60^{\circ}}{2}$

$45^{\circ} = \frac{D + 60^{\circ}}{2}$

$D = (90 - 60)^{\circ}$

$= 30^{\circ}$

The angle of a prism is 60∘. If the refractive index of the material of the prism is √2 the angle of minimum deviation is :45∘60∘30∘sin−1(23)

μ=sin(D+A2)sin(A/2) μ=2sin(D+602) =2sin(D+602) 1√2=sinD+60∘2 45∘=D+60∘2 D=(90−60)∘ =30∘

The angle of a prism is 60 $^{\circ}$ . If the refractive index of the material of the prism is $\sqrt{2}$ the angle of minimum deviation is :
45 $^{\circ}$
60 $^{\circ}$
30 $^{\circ}$
sin $^{- 1} (\frac{2}{3})$

$μ = \frac{s i n (\frac{D + A}{2})}{s i n (A / 2)}$

$μ = 2 s i n (\frac{D + 60}{2})$

$= 2 s i n (\frac{D + 60}{2})$

$\frac{1}{\sqrt{2}} = s i n \frac{D + 60^{\circ}}{2}$

$45^{\circ} = \frac{D + 60^{\circ}}{2}$

$D = (90 - 60)^{\circ}$

$= 30^{\circ}$