Derive the expression for refractive index of the material of the prism in terms of angle of the prism and angle of minimum deviation.

In the given diagram,\$OP\$ is the incidence ray, which is making the angle \$i_1\$ with normal, and \$QR\$ is the angle of emergence, which is represented by \$i_2\$. \$A\$ is the prism angle and \$\mu\$ is the refractive index of the prism.Now, We know that,\$A=\$Prism angle, \$\delta=\$Angle of deviation, \$i_1=\$Angle of incidence, \$i_2=\$Angle of emergent.In the case of minimum deviation,\$\angle r_1=\angle r_2=\angle r\$\$A=\angle r_1+\angle r_2\$SO, \$A=\angle r+\angle r=\angle 2r\$\$\angle r=\dfrac{A}{2}\$Now, again\$A+\delta=i_1+i_2\ \ {(\because}\$ In the case of minimum deviation \$i_1=i_2=i\$ and \$\delta=\delta_m)\$So, \$A+\delta_m=i+i=2i\$Now, \$i=\dfrac{A+\delta_m}{2}\$Now, from snell's rule,\$\mu=\dfrac{sin\ i}{sin\ r}\$\$\mu=\dfrac{sin \dfrac{A+\delta_m}{2}}{sin \dfrac{A}{2}}\$

A ray of light passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence and each of these equal to \$3/4\$ of the angle of the prism. The angle of deviation is

Given:Angle of incident \$i=\$ Angle of emergence \$=\dfrac34 A\$Where , \$A=\$ Angle of prismSince prism is equilateral \$A=60\$\$\therefore i=e= 60^\circ \times \dfrac34=45^\circ\$From prism formula:Angle of deviation , \$\delta = i+e-A= 45+45-60= 30^o\$

The refracting angle of a prism is A and the refractive index of the material of the prism is \$\cot{\dfrac{A}{2}}\$. The angle of minimum deviation of the prism is

It is given \$\mu =cot A/2\$We know that:\$\mu= \dfrac{sin \left ( \dfrac{\delta m+A}{2} \right )}{sin A/2}\$\$\Rightarrow cot A/2 =\dfrac{sin \left ( \dfrac{\delta m+A}{2} \right )}{sin \left ( A/2 \right )}\$\$\Rightarrow \dfrac{cos A/2}{sin A/2}= \dfrac{sin\left ( \dfrac{\delta m+A }{2}\right )}{sin A/2}\$\$\Rightarrow sin \left ( \pi /2-A/2 \right )=sin \left (\dfrac{ \delta m+A }{2}\right )\$\$\Rightarrow \pi /2-A/2=\dfrac{\delta m+A}{2}\$\$\Rightarrow \pi -2A=\delta m\$

A prism of refractive index \$\sqrt2\$ has refracting angle \$60^0\$. In order that a ray suffers minimum deviation it should be incident at an angle:

Given: A prism of refractive index \$\sqrt 2\$ has refracting angle \$60^\circ\$. To find the angle of incidence in order that a ray suffers minimum deviation Solution:As per the given criteria,refractive index of the prism, \$\mu=\sqrt 2\$Angle of the prism, \$A=60^\circ\$For minimum angle of deviation we have angle of incidence is equal to angle of emergence, i.e., \$i=e\$Hence, \$i=\dfrac {A+\delta_m}2\$, where \$\delta_m\$ is the minimum deviation angle.We know,\$\mu=\dfrac {\sin\left(\dfrac{A+\delta_m}{2}\right)}{\sin \dfrac A2}\\\implies \sqrt 2=\dfrac {\sin i}{\sin \dfrac {60}2}\\\implies \sin i=\sqrt 2\times \sin (30)\\\implies \sin i=\sqrt 2\times \dfrac12\$Multiply and divide by \$\sqrt 2\$, we get\$\sin i=\dfrac 1{\sqrt 2}\\\implies i=45^\circ\$In order that a ray suffers minimum deviation it should be incident at an angle \$45^\circ\$

If the critical angle for total internal reflection from a medium to vacuum is \$30^o\$. Then velocity of light in the medium is:

Critical angle for total internal reflection \$\theta_c = 30^o\$As we know, refractive index of air medium \$n_{air} = 1\$Refractive index of the medium \$n = \dfrac{1}{\sin\theta_c}\$\$\therefore\$ \$n = \dfrac{1}{\sin 30^o} = \dfrac{1}{0.5}\$\$\implies\$ \$n = 2\$Velocity of light in medium \$v = \dfrac{c}{n}\$\$\therefore\$ \$v = \dfrac{3\times 10^8}{2} = 1.5 \times 10^8 m/s\$

A ray is incident at an angle of incidence on one surface of a prism of small angle \$A\$ and emerges normally from the opposite surface. If the refractive index of the material of the prism is \$\mu\$, the angle of incidence \$i\$ is nearly equal to :

Hint: Here, we have to apply the concept of angle of deviationSolution:Step1: Draw free body diagram Step2: Apply given conditions According to free body diagram,i=Angle of incident, A=Small angle of prism & µ=Refractive index of material Assume, e= angle of emergence given that, the incident ray emerges normally from the opposite surface,Hence, \$\angle e = 0.....(1)\$ Step3: Find angle of deviationWe know the angle of deviation (\$\delta \$) can be,\$\delta = \left( {\mu - 1} \right)A.....(2)\$ In case of prism, we know that, sum of small angle of prism & angle of deviation in a prism is equal to the sum of angle of incidence and angle of emergence. So, we get, \$\delta + A = i + e.....(3)\$ Submit values from equation (1) & (2) in equation (3) we get, \$\left( {\mu - 1} \right)A + A = i + 0\$ \$ \Rightarrow \mu A - A + A = i\$\$ \Rightarrow \mu A = i\$ or \$i = \mu A\$ Hence, option (D) is correct.

For a glass prism \$(\mu = \sqrt{3})\$ the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

Using, \$\mu = \dfrac{sin\left ( \dfrac{A + \delta_m}{2} \right )}{sin \dfrac{A}{2}}\$According to question, \$\delta_m \,= \, A\$\$\therefore \sqrt{3}\, = \, \dfrac{sin\dfrac{A + A}{2}}{sin\dfrac{A}{2}}\$(Given \$\mu = \sqrt{3})\$\$\sqrt{3} = \dfrac{sinA}{sin\, \dfrac{A}{2}} = \dfrac{2sin\dfrac{A}{2}cos \, \dfrac{A}{2}}{sin\dfrac{A}{2}}\, or \, cos\dfrac{A}{2} = \dfrac{\sqrt{3}}{2} = cos 30^{\circ}\$\$\Rightarrow \dfrac{A}{2} \, = \, 30^{\circ} \Rightarrow A \, = \, 60^{\circ}\$

A ray of light is incident normally on one of the faces of a prism of apex angle \${30}^{o}\$ and refractive index \$\sqrt 2\$. The angle of deviation of the ray is

Using geometry in the figure, we get incident angle at prism face AC \$i = 30^o\$Snell's law at P : \$\mu\$ \$sin i = 1\times sinr\$\$\therefore\$ \$\sqrt{2} \times sin30^o = sinr \$ \$\implies r = sin^{-1} \bigg(\dfrac{1}{\sqrt{2}}\bigg) = 45^o\$Thus deviation of ray: \$\delta = r-i = 45^o -30^o =15^o\$

A ray of light is incident at an angle of 60\$^{\circ}\$on one of the faces of a prism which has an angle of 30\$^{\circ}\$. The ray emerging out of prism makes an angle of 30\$^{\circ}\$ with the incident ray. The refractive index of material is

\$\delta =30^{\circ} A=30^{\circ} \$ \$i_{1} =60^{\circ}\$\$\delta =\left ( i_{1}+i_{2} \right )-A\$\$30=60+i_{2}-30\$\$\Rightarrow i_{2} =0\$\$\Rightarrow r_{2}=0\$\$r_{1}=30\$\$\mu\times sin r_{1} =sin i\$\$\mu\times \dfrac{1}{2}=\dfrac{\sqrt{3}}{2} \$\$\mu =1.732\$

Problems on Tir in Prism | Definition, Examples, Diagrams

example

Total internal reflection in a Prism

Example: A ray

O P

monochromatic light is incident on the face

A B

of prism

A B C D

near vertex

B

at an incident angle of

6 0^{o}

(see figure). If the refractive index of the material of the prism is

3

, then find the angle between incident and emergent ray.

Solution:
Ray OP goes parallel to the side BC. Hence the angle of incidence of this ray on the side CD becomes

4 5^{\circ}

.Applying Snell's Law at the point Q,

3 sin 4 5^{\circ} = sin r > 1

Hence, this is a case of total internal reflection.Thus the angle of incidence on Q is equal to the angle of reflection.This forms

3 0^{\circ}

as angle of incidence on side AD.By Snell's Law,

3 sin 3 0^{\circ} = sin r = \frac{3}{2}

⟹ r = 6 0^{\circ}

Thus the angle between rays OP and RS is

9 0^{\circ}

example

Problem on TIR in prism

Problem:
A ray of light is incident normally on one face of a prism as shown in figure. The refractive index of the material of the prism is

\frac{5}{3}

and the prism is immersed in water of refractive index

\frac{4}{3}

then find angle of emergence and point of TIR.

Solution:
Initially at P, there is TIR, There force at Q (snells law)

\frac{5}{3} s i n 3 0^{0} = \frac{4}{3} s i n Θ_{2}

\Rightarrow s i n Θ_{2} = \frac{5}{8} \Rightarrow Θ_{2} = s i n^{- 1} (\frac{5}{8})

μ_{w} > \frac{5}{2 3} \Rightarrow s i n C > \frac{5 / 2 3}{2 / 3} \Rightarrow C > 6 0^{0}

Therefore, now TIR will not occur at P

example

Application of total internal reflection in a prism.

Many optical instruments use right -angled prisms to reflect a beam of light through $9 0^{o}$ or $18 0^{o}$ (By total internal reflection) such as cameras, binoculars, periscope and telescope.

When a ray of light strikes a face of prism perpendicular, it enters the prism without deviation and strikes the hypotenuse at an angle of $4 5^{o}$ . Since the angle of incidence $4 5^{o}$ is greater than critical angle of the glass which is $4 2^{o}$ , the light is totally reflected by the prism through an angle of $9 0^{o}$ . Two such prisms are used in periscope. The light is totally reflected by the prism by an angle of $18 0^{o}$ . Two such prisms are used in binoculars.

Prism is also used to invert images without changing their size.

Problems on Tir in Prism

example

Total internal reflection in a Prism

example

Problem on TIR in prism

example

Application of total internal reflection in a prism.

REVISE WITH CONCEPTS

Refraction Through Prism

Deviation Produced by a Triangular Prism

Relation Between Angle of Prism and Angle of Deviation, Refractive Index of Materials

Minimum Deviation in a Triangular Prism

Deviation in Reflection, Refraction and Prism

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