Question

Angle of prism is 60° and refractive index of prism is 1.414 find the minimum deviation

Answer 1

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Answer 2

Concept:

When a ray of light travels from one medium to another with a different refractive index, the angle of deviation is the angle that results from the difference between the angles of incidence and refraction.

The optical medium's refractive index is a dimensionless number that indicates how well the medium bends light.

Given:

The angle of the prism is $60^o$.

The refractive index of the prism is $1.414$ .

Find:

The minimum deviation of the ray of light.

Solution:

The angel of prism, $A=60^o$

The refractive index, $\mu=1.414$

The ray of light when incident on a prism and if the ray passes symmetrically.

Then, $r_1+r_2=A$

Now, the refractive index can be given as:

$\mu =\frac{\frac{sin(A+\delta _m)}{2}}{sin(30^o)}$

Where $\delta _m$ is the angle of deviation.

$1.414=\frac{\frac{sin(60+\delta _m)}{2}}{sin(30^o)}$

$\frac{sin(60+\delta _m)}{2}=0.707$

$\frac{(60^o+\delta _m)}{2}=45^o$

$\delta _m=30^o$

The minimum deviation is $30^o$.

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