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>> $Introduction to Prism$

Introduction to Prism

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: 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: Angle of minimum deviation for a prism of refractive index 1.5, is equal to the angle of the prism. Then the angle of the prism is : (Given \$cos 41^{\circ}=0.75\$)

Let \$\delta_m\$ be the angle of minimum deviation \$A=\delta_m\$\$A=\delta_m\$\$\mu=\dfrac{\sin \dfrac{(A+\delta_m)}{2}}{\sin \left(\dfrac{A}{2}\right)}\$\$1.5=\dfrac{\sin \left(\dfrac{A+A}{2}\right)}{\sin \left(\dfrac{A}{2}\right)}\$\$1.5=\dfrac{\sin 2\left(\dfrac{A}{2}\right)}{\sin \left(\dfrac{A}{2}\right)}=\dfrac{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}{\sin \left(\dfrac{A}{2}\right)}\$\$1.5=2\cos \dfrac{A}{2}\$\$0.75=\cos \dfrac{A}{2}\$\$\dfrac{A}{2}=\cos^{-1}(0.75)\$\$\dfrac{A}{2}=41^o\$\$\boxed{A=82^o}\$

Q: A ray of light passing through a prism having \$\mu = \sqrt 2\$ suffer minimum deviation. It is found that angle of incidence is double the angle of refraction within the prism. Find angle of the prism.

if \$\angle i = \$ incidence angle\$\angle r =\$ refracted angle\$\angle i = 2 \times \angle r\$\$\dfrac{\sin i}{\sin r} = \mu\$\$\dfrac{\sin 2r}{\sin r} = \dfrac{2 \sin r \cos r}{\sin r} = 2 \cos r = \sqrt{2}\$\$\cos r = \dfrac{1}{\sqrt{2}}\$\$r = 45^{\circ}\$For minimum deviation, \$r_1=r_2=r\$\$\therefore a=2\times r\$\$\angle a = 2 \times 45^{\circ}\$ \$= 90^{\circ}\$

Q: Copy the diagram given below and complete the path of light ray till it emerges out of the prism. The critical angle of glass is \$42^o\$. In your diagram mark the angles wherever necessary.

The diagram is shown. The ray \$PQ\$ enters the prism from the side \$AB\$, its angle of incidence is \$0°\$. Thus it does not undergo any refraction. Its angle of refraction is also \$0°\$. When this light ray reaches the base of the prism (BC), its angle of incidence (on the boundary \$BC\$) is \$60°\$ which is more than its critical angle (\$42°\$). Thus this incident light ray (\$QR\$) does not enter into the next medium, instead it undergoes total internal reflection. When this reflected light ray (\$RS\$) reaches the side \$AC\$, its angle of incidence is \$0°\$. Again, it does not undergo any refraction, which causes the angle of emergence to become \$0°\$.

Q: A ray of monochromatic light is incident on one refracting face of a prism of angle \$75^o\$. It passes through the prism and is incident on the other face at the critical angle. If the refractive index of the material of the prism is \$\sqrt 2\$, the angle of incidence on the first face of the prism is :

The correct option is BHint: Consider snells law and Critical angleCritical angle is given as,\$c={{\sin }^{-1}}\left( \dfrac{1}{\mu } \right) \$ Where, \$\mu \$ is the refractive index.In order to describe the relation between the angle of incidence and reflection snells law is used.\$\dfrac{{\sin i}}{{\sin r}} = \mu \$Where \$i\$ is the angle of incidence and \$r\$ is the angle of reflection.Explanation for correct answer:Step 1: Information requiredCritical angle is given as,\$c={{\sin }^{-1}}\left( \dfrac{1}{\mu } \right) \$ Where, \$\mu \$ is the refractive index.In order to describe the relation between the angle of incidence and reflection snells law is used.\$\dfrac{{\sin i}}{{\sin r}} = \mu \$Here, \$i\$ is angle of incidence \$r\$ is the angle of reflection.Angle of prism \$A = {75^o}\$ Refractive index \$\mu = \sqrt 2 \$ Step 2: Calculate the angle of reflection,The angle of the prism is the sum of reflection and critical angle.Then,\$A = r + c\$ Substitute the value of critical angle and angle of prism in the above equation \$A=r+{{\sin }^{-1}}\left( \dfrac{1}{\mu } \right) \$ \$75=r+{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right) \$\$75 = r + 45\$\$r = 75 - 45\$ \$r = {45^o}\$ So, the angle of reflection is \$r = {45^o}\$.Step 3: Calculate angle of incidence,Consider snells law,\$\dfrac{{\sin i}}{{\sin r}} = \mu\$\$\dfrac{{\sin i}}{{\sin 30}} = \sqrt 2\$\$\sin i = \sqrt 2 \times \dfrac{1}{2}\$\$\sin i = \dfrac{1}{{\sqrt 2 }}\$\$i = {\sin ^{ - 1}}\dfrac{1}{{\sqrt 2 }}\$\$i = \,{45^o}\$Hence the angle of incidence is \$i = \,{45^o}\$.

Q: Two vertical plane mirrors are inclined at an angle of \$60^0\$ with each other. A ray of light travelling horizontally is reflected first from one mirror and then from the other. The resultant deviation is

\$\begin{array}{l} From\, the\, figure \\ a+b={ 90^{ 0 } } \\ \Rightarrow b={ 90^{ 0 } }-a \\ Now, \\ b+60+c=180 \\ \Rightarrow c=180-60-b \\ =180-\left( { 90-a } \right) \\ =30+a \\ Now, \\ d+c=90 \\ d=90-c \\ =90-\left( { 30+a } \right) \\ =60-a \\ Thus,\, in\, triangle\, PQR \\ \angle Q=2a \\ \angle R=2d=2\left( { 60-a } \right) =120-2a \\ so,\, angle\, \angle QPT\, is\, =\angle Q+\angle R=120 \\ Now, \\ angle\, of\, deviation\, =reflex\, \angle QPT \\ =360-120 \\ ={ 240^{ \circ } } \\ Hence,\, the\, option\, D\, is\, the\, correct\, answer. \end{array}\$

Q: Draw a ray diagram to show the angle of deviation when a ray of light passes through a glass prism.

Angle of Deviation is defined as the angle between the incident and the emergent ray. As the light enters the first refracting surface of the prism it gets refracted and is incident at the second refracting surface and emerges out at an angle of emergence from the prism. The angle of deviation of is the angle of diversion which is produced in the path of the incident light ray striking at the prism

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