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\$

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>>Class 12
>>Physics
>>Ray Optics and Optical Instruments
>>Prism
>> $Deviation in Reflection, Re...$

Deviation in Reflection, Refraction and Prism

Q: 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}}\$

Q: 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\$

Q: 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\$

Q: 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\$

Q: 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.

Q: The angle of incidence for a ray of light at a refracting surface of a prism is \${45}^{o}\$. The angle of prism is \${60}^{o}\$. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are

Angle of incidence, \$i=45^0\$Angle of prism, \$A=60^0\$Angle of minimum deviation, \$\delta_m =2i-A=30^0\$\$\Rightarrow \mu=\dfrac{sin \left ( \dfrac{A+\delta_m}{2}\right )}{sin \dfrac{A}{2}}=\dfrac{sin\ 45^0}{sin\ 30^0}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\$

Q: 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\$

Q: 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\$

Q: For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index:

The angle of minimum deviation is given as\$\delta_{min} = i + e-A\$for minimum deviation\$\delta_{min}= A\$ then\$2A = i + e\$in case of \$\delta_{min};\ i = e\$\$2A = 2i\ r_{1} = r_{2} = \dfrac {A}{2}\$\$i = A = 90^{\circ}\$\$1\sin r = n\sin r_{1}\$\$\sin A = n\sin \dfrac {A}{2}\$\$2\sin \dfrac {A}{2} \cos \dfrac {A}{2} = n\sin \dfrac {A}{2}\$\$2\cos \dfrac {A}{2} = n\$when \$A = 90^{\circ} = i_{min}\$then \$n_{min} = \sqrt {2}\$\$i = A = 0\ \$\$n_{max} = 2\$.

Physics

diagram

Plot of Angle of Deviation Vs Angle of Prism

Above plot show variation of angle of deviation with angle of incidence for prism angle 60 degree and refractive index 1.5.

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Refraction Through Prism

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Minimum Deviation in a Triangular Prism

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Problems on Tir in Prism

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