Consider an optical system placed in water \left ( \mu=4/3 \right ) PQRS is a hollow glass slab filled with air \left ( \mu=1 \right ) This slab is kept on the principle axis of a lens in such a way that surface PS makes an angle \theta with the axis of the lens. \left ( 0<\theta < 180^{\circ} \right ) , then

if \theta \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} Similarly if \theta > 90^{\circ} , thenfor No TIR \theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4} so \left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4}<\theta <\dfrac{\pi }{2}+sin ^{-1}\dfrac{3}{4} \right ) For calculating focal length of lens in air \dfrac{1}{f}=\left ( \mu_{rel}-1 \right )\left ( \dfrac{1}{R_{1}} -\dfrac{1}{R_{2}}\right ) \dfrac{1}{50}=\left ( \dfrac{3}{2}-1 \right )\left ( k \right )\Rightarrow k=\dfrac{1}{25} f_{w}^{g}=4f_{a}^{G}=4 \times 50cm =200m \dfrac{1}{f_{w}^{o}}=\left ( \dfrac{3}{2}\dfrac{3}{4}-1 \right )\left ( k \right )=\dfrac{k}{8} \Rightarrow f_{w}=200cm So that final image will be formed 200 cm in right of lens. \Rightarrow Image \ distance,OI = (100 + 200) cm = 300 cm Ans. (a) and (b) \left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} <\theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4}\right ) (d) OI = (100 + 200) cm = 300 cm but this distance is valid only when \theta is between \theta _{min} to \theta _{max} Ans. (a) and (b) \left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} <\theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4}\right )

Ray Optics And Optical Instruments

Q: An object starts moving at an angle at 45^o with the principal axis as shown in fig in front of a biconvex lens of focal length +10\ cm . If \theta denotes the angle at which image starts to move with principal axis, then:

u= -20 ; f= +10 So, from \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} , we get: v= +20 Magnification: \dfrac{v}{u}= -1 ............ (i)Hence vertical velocity of image is same as that of object but in reverse direction.Let the horizontal velocity be V_{Ix} We have: v=\dfrac{fu}{u-f} Differentiating w.r.t. time, we get: V_{Ix}=\dfrac{f^{2}}{(u-f)^{2}} \times V_{object} \Rightarrow V_{Ix}=V_{object} [using (i)]Hence angle made by velocity vector of image is -\dfrac{\pi}{4}

Q: Two identical glass rods S_{1} and S_{2} (refractive index = 1.5) have one convex end of radius of curvature 10\ cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S_{1} on its axis at a distance of 50\ cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S_{2}. The distance d is:

Applying \left( \dfrac { { \mu }_{ 2 } }{ v } -\dfrac { { \mu }_{ 1 } }{ u } =\dfrac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ R } \right) two times, we will get \Rightarrow \dfrac{1}{v}-\dfrac{1.5}{-50}=\dfrac{1-1.5}{-10} \Rightarrow \dfrac{1}{v}=\dfrac{0.5}{10}-\dfrac{1.5}{50}=\dfrac{2.5-1.5}{50} \Rightarrow v=50. \Rightarrow AB=d, AI_1=50cm BI_1=\left(d-50\right)cm Again, \dfrac{1.5}{\infty}-\dfrac{1}{-\left(d-50\right)}=\dfrac{1.5-1}{10} \Rightarrow \dfrac{1}{d-50}=\dfrac{1}{20} \Rightarrow d=70\;cm. Hence, the answer is 70\;cm.

Q: A light ray 1 is incident on a plane mirror M . The mirror is rotated in the direction as shown in the figure by an arrow at frequency \dfrac{9}{\pi} rev/s . The light reflected by the mirror is received on the wall W at a distance 10m from the axis of rotation. When the angle of incidence becomes {37}^{o} the speed of the spot (a point) on the wall is:

Rotational speed of reflected light is given by, \omega' = 2\times \dfrac{9}{\pi} \times 2\pi = 18 \ rad/s Now, v' = R\omega' The velocity of spot on wall is given as, v \cos (90^o - 37^o) = R\omega' v \cos 53^o = R\omega' v = \dfrac{10}{\cos 53^o} \times \dfrac{36}{\cos 53^o} = \dfrac{36\times 25}{9} = 1000 rad/s

Q: 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 \dfrac{5}{3} and the prism is immersed in water of refractive index \dfrac{4}{3} then

Initially at P, there is TIR, There force at Q (snells law) \dfrac{5}{3} sin 30^{0} = \dfrac{4}{3} sin \Theta_{2} \Rightarrow sin \Theta_{2} = \dfrac{5}{8} \Rightarrow \Theta_{2} = sin^{-1}(\dfrac{5}{8}) If \mu_{w} > \dfrac{5}{2 \sqrt{3}} \Rightarrow sin C > \dfrac{5/2\sqrt{3}}{2/3} \Rightarrow C > 60^{0} Therefore, now TIR will not occur at P

Q: Consider an optical system placed in water \left ( \mu=4/3 \right ) PQRS is a hollow glass slab filled with air \left ( \mu=1 \right ) This slab is kept on the principle axis of a lens in such a way that surface PS makes an angle \theta with the axis of the lens. \left ( 0<\theta < 180^{\circ} \right ) , then

if \theta \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} Similarly if \theta > 90^{\circ} , thenfor No TIR \theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4} so \left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4}<\theta <\dfrac{\pi }{2}+sin ^{-1}\dfrac{3}{4} \right ) For calculating focal length of lens in air \dfrac{1}{f}=\left ( \mu_{rel}-1 \right )\left ( \dfrac{1}{R_{1}} -\dfrac{1}{R_{2}}\right ) \dfrac{1}{50}=\left ( \dfrac{3}{2}-1 \right )\left ( k \right )\Rightarrow k=\dfrac{1}{25} f_{w}^{g}=4f_{a}^{G}=4 \times 50cm =200m \dfrac{1}{f_{w}^{o}}=\left ( \dfrac{3}{2}\dfrac{3}{4}-1 \right )\left ( k \right )=\dfrac{k}{8} \Rightarrow f_{w}=200cm So that final image will be formed 200 cm in right of lens. \Rightarrow Image \ distance,OI = (100 + 200) cm = 300 cm Ans. (a) and (b) \left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} <\theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4}\right ) (d) OI = (100 + 200) cm = 300 cm but this distance is valid only when \theta is between \theta _{min} to \theta _{max} Ans. (a) and (b) \left ( \dfrac{\pi }{2}- sin ^{-1}\dfrac{3}{4} <\theta <\dfrac{\pi }{2}+ sin ^{-1}\dfrac{3}{4}\right )

Q: Two parallel rays are incident on a thin lens of refractive index 1.5 as shown in figure and radius of curvature of its surfaces are 30 cm & 60 cm . Origin of coordinate axes is at the optical centre of the lens.Then

We use the lens maker's formula, \dfrac{1}{f} = (\mu - 1)(\dfrac{1}{{R}_{1}} - \dfrac{1}{{R}_{2}}) \mu = 1.5 , {R}_{1} = -60, {R}_{2} = -30 f = 120 Since, a lens converges a parallel beam of light on its focus, the y coordinate of the lens is, y = -f \theta y = 120 \times \dfrac{6 \pi}{180} Thus, y = -4 \pi

Q: A thin biconvex lens of refractive index 3/2 and radius of curvature 50\ cm is placed on a reflecting convex surface of radius of curvature 100\ cm . A point object is placed on the principal axis of the system such that its final image coincides with itself. Now, few drops of a transparent liquid is placed between the mirror and lens such that final image of the object is at infinity. Find refractive index of the liquid used. Also, find the position of the object.

Given: A thin biconvex lens of refractive index 3/2 and radius of curvature 50 cm is placed on a reflecting convex surface of radius of curvature 100 cm. A point object is placed on the principal axis of the system such that its final image coincides with itself. Now, few drops of a transparent liquid is placed between the mirror and lens such that final image of the object is at infinity. To find refractive index of the liquid used. Also, find the position of the objectSolution:The system behaves as mirrorPower of lens, P_l=\dfrac 1{f_l}=\left(\dfrac 32-1\right)\left(\dfrac 1{50}-\dfrac 1{-50}\right)=\dfrac 1{50} Therefore power of equivalent mirror P_{EM}=2P_l+P_m=\dfrac 2{50}-\dfrac 1{50}=\dfrac 1{50} i.e., Focal length of equivalent mirror, f_{cm}=-50 Since image and the object coincides, hence position of the object is at the center of curvature of the equivalent mirror, i.e., at 2f_{cm}=100cm from lensWhen the liquid of refractive index be is placed between the lens and convex mirror, if forms a liquid lens of power, P_u=\left(\dfrac 1{-50}-\dfrac 1{100}\right)=\dfrac {3(1-\mu)}{100} Now, power of equivalent mirror 2P_l+2P_u+P_m\\\implies =\dfrac 2{50}+\dfrac {2\times 3(1-\mu)}{100}+\dfrac 1{-50}=\dfrac {3(1-\mu)+1}{50}...........(i) Since final image is at infinity, hence the object should be at its focal point i.e., f'_{EM}=100\\\implies \dfrac 1{f'_{EM}}=\dfrac 1{100}...........(ii) From eqn(i) and (ii), we get \dfrac {3(1-\mu)+1}{50}=\dfrac 1{100}\\\implies 6(1-\mu)+2=1\\\implies 6-6\mu=-1\\\implies \mu=\dfrac 76 is the required refractive index of the liquid used.

Q: A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown. The space between the lens and the mirror is then filled with water of refractive index 4/3 . It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own images. On representing with another liquid, the object and the image again coincide at a distance 25 cm from the lens. Calculate x , if the refractive index of the liquid is \dfrac{10+x}{10}

Let f_1 be the focal length of convex lens; radius of curvature of each curved face is R. \dfrac {1}{f_1}=(m-1)\left \{\dfrac {1}{R}-\left (\dfrac {1}{-R}\right )\right \}=(\mu-1)\dfrac {2}{R} \Rightarrow f_1=\dfrac {R}{2(\mu-1)}=\dfrac {R}{2\left (\dfrac {3}{2}-1\right )}=R When the space between the lens and mirror is filled by water of refractive index \mu_1=4/3 , then the focal length of liquid concave lens f_2 is \dfrac {1}{f_2}=(\mu_1-1)\left (-\dfrac {1}{R}-\infty \right ) \Rightarrow f_2=\dfrac {-R}{\mu_1-1}=-\dfrac {R}{\left (\dfrac {4}{3}-1\right )}=-3R The combined focal length of lenses is F_1=15 cm \therefore \dfrac {1}{F_1}=\dfrac {1}{f_1}+\dfrac {1}{f_2}\Rightarrow \dfrac {1}{15}=\dfrac {1}{R}-\dfrac {1}{3R}=\dfrac {3-1}{3R} \Rightarrow 3R=30\Rightarrow R=10 cm In the second case, F_2=25 cm . Let \mu_1=\mu_2 \therefore \dfrac {1}{F_2}=\dfrac {1}{f_1}+\dfrac {1}{f_2}\Rightarrow \dfrac {1}{25}=\dfrac {1}{10}+\dfrac {1}{f_2'} \Rightarrow \dfrac {1}{f'_2}=\dfrac {1}{25}-\dfrac {1}{10}=\dfrac {2-5}{50} \therefore f'_2=\dfrac {-50}{3}cm f'_2=\dfrac {R}{\mu_2-1}\Rightarrow \mu_2-1=-\dfrac {R}{f'_2}=\dfrac {-10}{(-50/3)}=\dfrac {3}{5}=0.6 \Rightarrow \mu_2=1+0.6=1.6

Q: AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence (in degrees) at Q so that the ray is totally reflected by both the slabs.

Let the minimum angle be \alpha For first surface to completely reflect, 2Sin \alpha \geqslant \sqrt{2} Sin 90 => sin \alpha \geqslant 1/\sqrt{2} => \alpha \geqslant 45^\circ For second surface to completely reflect, 2Sin \alpha \geqslant \sqrt{3} Sin 90 => sin \alpha \geqslant \sqrt{3}/2 => So satisfies both conditionAnswer. Minimum angle of incidence so as to totally reflect from both surface is \alpha = 60 ^\circ

Q: A composite slab consisting of different media is placed in front of concave mirror of radius of curvature 150\ cm as shown in Fig. The whole arrangement is immersed in water. Locate the final image of point object O .

Given: A composite slab consisting of different media is placed in front of a concave mirror of radius of curvature 150 cm. The whole arrangement is immersed in water.To find the location of the final image of point object O.Solution:The image formed after 4 refraction would be taken as the object. The distance of object (4 ^th image) from mirror's pole is given by x=\left(\dfrac {20}{\dfrac 43}+\dfrac {54}{51}+\dfrac {24}{1.0}+\dfrac {45}{1.5}\right)\times\dfrac 43+10\\\implies x=(15+1.06+24+30)\times 1.3+10\\\implies x= 103cm Hence, the ray will retrace its path,i.e., it is formed on 54cm slab.