The angle of minimum deviation produced by a $60^{\circ}$ prism is $40^{\circ} .$ Calculate the refractive index of the material of the prism.
Answer
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Hint: We know that reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. The wavelength decreases as the light enters the medium and the light wave changes direction.
Complete step by step answer
We know that the incident ray, reflected ray and the normal, to the interface of any two given mediums; all lie in the same plane. The ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.
Minimum deviation $\delta=2 \mathrm{i}-\mathrm{A}$
where A is prism angle $A=60^{\circ}$
So $\delta=40^{\circ}$
we know
$\delta=2 \mathrm{i}-\mathrm{A}$
$40=2 \mathrm{i}-60$
$\mathrm{i}=50^{\circ}$
For minimum angle the angle of refraction should be half of prism angle.
So $\mathrm{r}=\mathrm{A} / 2=30^{\circ}$
Now using Snell's law which states that the product of the index of refraction by the sine of the angle of incidence is constant for any ray of light striking the separating surface of two media.
So, we can write that:
$\mathrm{n}_{\text {air }} \operatorname{sini}=\mathrm{n}_{\text {glass }} \sin \mathrm{r}$
$\sin 50=\mathrm{n}_{\mathrm{glass}} \sin 30$
$\mathrm{n}_{\mathrm{glass}}=1.53$
Note: We know that the smallest angle through which light is bent by an optical element or system. In a prism, the angle of deviation is a minimum if the incident and exiting rays form equal angles with the prism faces. The angle is important relative to prism spectroscopes because it can be easily determined. White light may be separated into its spectral colours by dispersion in a prism. Prisms are typically characterized by their angle of minimum deviation d. This minimum deviation is achieved by adjusting the incident angle until the ray passes through the prism parallel to the bottom of the prism.
Complete step by step answer
We know that the incident ray, reflected ray and the normal, to the interface of any two given mediums; all lie in the same plane. The ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.
Minimum deviation $\delta=2 \mathrm{i}-\mathrm{A}$
where A is prism angle $A=60^{\circ}$
So $\delta=40^{\circ}$
we know
$\delta=2 \mathrm{i}-\mathrm{A}$
$40=2 \mathrm{i}-60$
$\mathrm{i}=50^{\circ}$
For minimum angle the angle of refraction should be half of prism angle.
So $\mathrm{r}=\mathrm{A} / 2=30^{\circ}$
Now using Snell's law which states that the product of the index of refraction by the sine of the angle of incidence is constant for any ray of light striking the separating surface of two media.
So, we can write that:
$\mathrm{n}_{\text {air }} \operatorname{sini}=\mathrm{n}_{\text {glass }} \sin \mathrm{r}$
$\sin 50=\mathrm{n}_{\mathrm{glass}} \sin 30$
$\mathrm{n}_{\mathrm{glass}}=1.53$
Note: We know that the smallest angle through which light is bent by an optical element or system. In a prism, the angle of deviation is a minimum if the incident and exiting rays form equal angles with the prism faces. The angle is important relative to prism spectroscopes because it can be easily determined. White light may be separated into its spectral colours by dispersion in a prism. Prisms are typically characterized by their angle of minimum deviation d. This minimum deviation is achieved by adjusting the incident angle until the ray passes through the prism parallel to the bottom of the prism.
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