(A) For a thin prism,the angle of deviation $D$ is given by the formula $D = (\mu - 1) A$,where $\mu$ is the refractive index and $A$ is the prism ang

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(A) For a thin prism,the angle of deviation $D$ is given by the formula $D = (\mu - 1) A$,where $\mu$ is the refractive index and $A$ is the prism angle.Since the refractive index for blue light $(\mu_{blue} = 1.525)$ is greater than the refractive index for red light $(\mu_{red} = 1.520)$,we have $\mu_{blue} > \mu_{red}$.Substituting these into the deviation formula,we get $D_{blue} > D_{red}$.Given that $D_1$ is the deviation for red light and $D_2$ is the deviation for blue light,it follows that $D_2 > D_1$ or $D_1 < D_2$.

Class 12 Physics Ray Optics and Optical Instruments Refraction Through Prism

Easy

The refractive index of glass for red light is $1.520$ and for blue light is $1.525$. Let $D_1$ and $D_2$ be the angles of deviation for red and blue light,respectively,in a prism of this glass. Then:

A
$D_1 < D_2$
B
$D_1 = D_2$
C
$D_1$ and $D_2$ depend on the angle of the prism.
D
$D_1 > D_2$

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Refraction Through Prism

$A$ beam of monochromatic light is incident at $i = 50^{\circ}$ on one face of an equilateral prism. If the angle of emergence is $e = 40^{\circ}$,then the angle of minimum deviation $\delta_m$ is:

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For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

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$A$ small angled prism of refractive index $1.6$ gives a deviation of $3.6^{\circ}$. The angle of prism is . . . . . . . (in $^{\circ}$)

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$A$ certain prism is found to produce a minimum deviation of $38^\circ$. It produces a deviation of $44^\circ$ when the angle of incidence is either $42^\circ$ or $62^\circ$. What is the angle of incidence when it is undergoing minimum deviation (in $^\circ$)?

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$A$ ray of light is incident at $60^{\circ}$ on one face of a prism of angle $30^{\circ}$ and the emergent ray makes $30^{\circ}$ with the incident ray. The refractive index of the prism is $(\sin 30^{\circ}=0.5, \sin 60^{\circ}=\sqrt{3}/2)$.

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The refractive index of glass for red light is $1.520$ and for blue light is $1.525$. Let $D_1$ and $D_2$ be the angles of deviation for red and blue light,respectively,in a prism of this glass. Then:

Similar Questions

$A$ beam of monochromatic light is incident at $i = 50^{\circ}$ on one face of an equilateral prism. If the angle of emergence is $e = 40^{\circ}$,then the angle of minimum deviation $\delta_m$ is:

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index $n$ of the prism satisfies.

$A$ small angled prism of refractive index $1.6$ gives a deviation of $3.6^{\circ}$. The angle of prism is . . . . . . . (in $^{\circ}$)

$A$ certain prism is found to produce a minimum deviation of $38^\circ$. It produces a deviation of $44^\circ$ when the angle of incidence is either $42^\circ$ or $62^\circ$. What is the angle of incidence when it is undergoing minimum deviation (in $^\circ$)?

$A$ ray of light is incident at $60^{\circ}$ on one face of a prism of angle $30^{\circ}$ and the emergent ray makes $30^{\circ}$ with the incident ray. The refractive index of the prism is $(\sin 30^{\circ}=0.5, \sin 60^{\circ}=\sqrt{3}/2)$.

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