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: Light enters from air to glass having refractive index \$1.50\$. What is the speed of light in the glass? The speed of light in vacuum is \$3 \times 10^{8}\ m\ s^{-1}\$.

Given that, Refractive index \$\mu =1.50\$ Speed of light \$c=3\times {{10}^{8}}\,m/s\$ We know that, \$ {{n}_{g}}=\dfrac{c}{{{v}_{g}}} \$ \$ {{v}_{g}}=\dfrac{3\times {{10}^{8}}}{1.50} \$ \$ {{v}_{g}}=2\times {{10}^{8}}\,m/s \$ Hence, the speed of light in glass is \$2\times {{10}^{8}}\,m/s\$

Q: The absolute refractive index of glass and water are \$3/2\$ and \$4/3\$, respectively. If the speed of light in glass is \$2 \times 10^8\$ m/s, calculate the speed of light in :(a) vacuum(b) water

Given : \$n_{water} = \dfrac{4}{3}\$ \$n_{glass} = \dfrac{3}{2}\$ \$v_{glass} = 2\times 10^8\$ m/s \$n_{air} = 1\$Speed of light in a medium \$v_{medium} = \dfrac{c}{n}\$ where \$c\$ and \$n\$ are the speed of light in vacuum and refractive index of medium respectively.(a) : \$v_{glass} = \dfrac{c}{n_{glass}}\$\$\therefore\$ \$2\times 10^8 = \dfrac{c}{(3/2)}\$ \$\implies c = 3\times 10^8\$ m/s(b) : \$v_{water} = \dfrac{c}{n_{water}}\$\$\therefore\$ \$v_{water} = \dfrac{3\times 10^8}{(4/3)} = 2.26\times 10^8\$ m/s

Q: A \$4.5\ cm\$ needle is placed \$12\ cm\$ away from a convex mirror of focal length \$15\ cm\$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Given: Height of the needle, \$h_1 = 4.5 cm\$.Object distance, \$u = −12 cm\$.The focal length of the convex mirror, \$f=15cm\$.Image distance, \$ v\$ The value of \$v\$ can be obtained using the mirror formula.\$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\$\$\dfrac{1}{v}+\dfrac{1}{-12}=\dfrac{1}{15}\$\$\dfrac{1}{v} = \dfrac{1}{12}+\dfrac{1}{15}\$\$\dfrac{1}{v} = \dfrac{9}{60}\$\$\therefore v \approx 6.7cm\$Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.The image size is given by the magnification formula.\$m=\dfrac{h^{'}}{h}=-\dfrac{v}{u}\$\$h'=\dfrac{6.7\times4.5}{12}\$\$\Rightarrow h'=+2.5cm\$So, \$m=\dfrac{2.5}{4.5}\$\$m=0.56\$The height of the image is \$2.5 cm\$. The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the size of the image will reduce gradually.

Q: A Concave mirror produces three times magnified real image of an object placed at \$10cm\$ infront of it . Where is the image located?

Given magnification \$m=3\$and object distance \$u=10\$ cm and we know,Magnification \$m=\dfrac{-v}{u}\$\$-v=3\times -10\$\$v=30\$ cmThus the image is formed at distance of 30 cm.

Question 1

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

Accepted Answer

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}}$

Question 2

Light enters from air to glass having refractive index $1.50$. What is the speed of light in the glass? The speed of light in vacuum is $3 \times 10^{8}\ m\ s^{-1}$.

Accepted Answer

Given that,&#10;&#10;Refractive index $\mu =1.50$ &#10;&#10;Speed of light $c=3\times {{10}^{8}}\,m/s$ &#10;&#10;We know that,&#10;&#10;  $ {{n}_{g}}=\dfrac{c}{{{v}_{g}}}&#10;$ &#10;&#10; $ {{v}_{g}}=\dfrac{3\times&#10;{{10}^{8}}}{1.50} $ &#10;&#10; $ {{v}_{g}}=2\times&#10;{{10}^{8}}\,m/s $ &#10;&#10;Hence, the speed of light in glass is $2\times&#10;{{10}^{8}}\,m/s$

Question 3

The absolute refractive index of glass and water are $3/2$ and $4/3$, respectively. If the speed of light in glass is $2 \times 10^8$ m/s, calculate the speed of light in :(a) vacuum(b) water

Accepted Answer

Given :    $n_{water} = \dfrac{4}{3}$           $n_{glass} = \dfrac{3}{2}$             $v_{glass} = 2\times 10^8$  m/s            $n_{air} = 1$Speed of light in a medium        $v_{medium} = \dfrac{c}{n}$         where  $c$ and $n$ are the speed of light in vacuum and refractive index of medium respectively.(a) :     $v_{glass} = \dfrac{c}{n_{glass}}$$\therefore$     $2\times 10^8  = \dfrac{c}{(3/2)}$                        $\implies c  = 3\times 10^8$  m/s(b) :     $v_{water} = \dfrac{c}{n_{water}}$$\therefore$     $v_{water}  = \dfrac{3\times 10^8}{(4/3)}  = 2.26\times 10^8$  m/s

Question 4

Why do stars twinkle? Explain.

Accepted Answer

The twinkling of stars is due to atmospheric refraction of star-light. The atmosphere is made of several layers and the refractive indices which keep on changing continuously due to this path of light rays from the star keep on changing their path continuously. As a consequence the number of rays entering, the pupil of the eye goes on changing with time and the stars appear twinkling.

Question 5

A $4.5\ cm$ needle is placed $12\ cm$ away from a convex mirror of focal length $15\ cm$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Accepted Answer

Given: Height of the needle, $h_1 = 4.5 cm$.Object distance, $u = &#8722;12 cm$.The focal length of the convex mirror, $f=15cm$.Image distance, $ v$ The value of $v$ can be obtained using the mirror formula.$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$\dfrac{1}{v}+\dfrac{1}{-12}=\dfrac{1}{15}$$\dfrac{1}{v} = \dfrac{1}{12}+\dfrac{1}{15}$$\dfrac{1}{v} = \dfrac{9}{60}$$\therefore v \approx 6.7cm$Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other&#10;side of the mirror.The image size is given by the magnification formula.$m=\dfrac{h^{'}}{h}=-\dfrac{v}{u}$$h'=\dfrac{6.7\times4.5}{12}$$\Rightarrow h'=+2.5cm$So, $m=\dfrac{2.5}{4.5}$$m=0.56$The height of the image is $2.5 cm$. The positive sign indicates that the image is erect,&#10;virtual, and diminished.&#10;If the needle is moved farther from the mirror, the size of the image will reduce gradually.

Question 6

A Concave mirror produces three times magnified real image of an object placed at $10cm$ infront of it . Where is the image located?

Accepted Answer

Given magnification $m=3$and object distance $u=10$ cm and we know,Magnification $m=\dfrac{-v}{u}$$-v=3\times -10$$v=30$ cmThus the image is formed at distance of 30 cm.

Question 7

At noon the sun appears white because:

Accepted Answer

At noon because the sun is overhead, the light is scattered the least and hence appears white. When it is overhead, it has lesser air to travel through and the scattering from dust and other particles is reduced if the distance to be travelled in air is reduced.

Question 8

State the characteristics of the image formed by a plane mirror.

Accepted Answer

(i) Plane mirror forms an erect image.(ii) It forms a virtual image.(iii) Size of the image is same as that of the object.(iv) Image is formed at the same distance behind the mirror as the object stands in front of it.(v)Image formed is a laterally inverted image i.e., right hand side of the object seems to be the left hand side and vice-versa.

Question 9

A spherical mirror and a thin spherical lens have each a focal length of $-15$ cm. The mirror and the lens are likely to be

Accepted Answer

By convention, the focal length of a concave mirror and a concave lens are taken as negative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.

Question 10

Which of the following statements is correct regarding rods and cones in the human eye?

Accepted Answer

The retina is the back part of the eye that contains the cells that respond to light. These specialized cells are called photoreceptors. There are 2 types of photoreceptors in the retina: rods and cones.The rods are most sensitive to light intensity changes. They contain only one type of light-sensitive pigment. Rods are not good for color vision. In a dim room, however, we use mainly our rods, but we are color blind. Rods are more numerous than cones in the periphery of the retina. There are about 120 million rods in the human retina.The cones are less sensitive to light. However, cones are more sensitive to one of three different colors (green, red or blue). Signals from the cones are sent to the brain which then translates these messages into the perception of color. Cones, however, work only in bright light.