A ray of light from a liquid \$\left ( \mu =\sqrt{3} \right )\$ is incident on a system of two right angled prism of refractive indices \$\sqrt{3} and \sqrt{2}\$ as shown in the figure. The light suffers zero net deviation when it emerges into air from surface CD. If the angle of incidence (in degrees) is \$5{n}\$. Find n ?

First, we apply the law of refraction for the interface AC,The incident angle will be \$90 - i\$\$ \sqrt{3}sin (90 - i) = \sqrt{2} sin r \$\$ \sqrt{3} cos i = \sqrt{2} sin r \$Now, for the interface CD,\$ \sqrt{2} sin (90 - r) = sin i \$Thus, \$ \sqrt{2} cos r = sin i \$We have two equations in i and r, squaring and adding these 2,we get,\$ 2 {cos}^{2} i + 1 = 2 \$This gives, \$cos \ i = \dfrac{1}{\sqrt{2}} \$Thus, \$i = 45\$

The refracting angle of a prism \$60^{\mathrm{o}}\$.The refractive index of the material of the prism is \$\sqrt{\dfrac{7}{3}}\$. The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

\$\begin{array}{l}\text { Given } A=60^{\circ}, \mu=\sqrt{\frac{7}{3}} \\\text { we know that } \gamma_{1}+r_{2}=A \\\Rightarrow r_{1}+r_{2}=60^{\circ}\end{array}\$\$\begin{array}{l}\text { Snell's law at first refraction } \\\qquad \begin{array}{l}\sin i=\mu \sin \gamma_{1} \\\operatorname{Sin}\gamma_{1}=\sqrt{\frac{3}{7}} \sin i\end{array} \\\text { Snell's law at second refraction }\end{array}\$\$\begin{aligned}\mu \sin r_{2} &=1 \\ & \Rightarrow \sin r_{2}=\sqrt{\frac{3}{7}} \\ r_{2}=\sin ^{-1}\left(\sqrt{\frac{3}{7}}\right) \\ r_{2}=& 40.89^{\circ} \Rightarrow r_{1}=19.1066^{\circ}\end{aligned}\$\$\begin{aligned}\Rightarrow\operatorname{Sin} r_{1} &=0.327 \\\sin i &=\sqrt{\frac{7}{3}} \sin r_{1} \\\sin i &=1 / 2 \\ i &=30^{\circ}\end{aligned} \$

A ray of light enters at grazing angle of incidence into an assembly of three isosceles right-angled prisms having refractive indices \$\mu_{1}=\sqrt{2}, \mu_{2}=\sqrt{x}\$ and \$\mu_{3}=\sqrt{3}\$ . If finally emergent light ray also emerges at grazing angle then calculate x :

Apply Snell's law on various surfaces one by one :1 sin \$90^{\circ}\$ \$=\mu_{1}\$ sin \$r_{1} \Rightarrow\$ sin \$r_{1} = \dfrac{1}{\sqrt{2}} \Rightarrow r_{1}=45^{\circ}\$ We have the incident angle for the second interface given using the relation sin\$i_2\$=cos\$r_1\$Thus we get\$\mu_{1}\$ cos \$r_{1}\$= \$\mu_{2}\$ sin \$r_{2} \Rightarrow \$ sin \$r_{2}\$ = \$\dfrac{1}{\mu_{2}}\$(cos\$r_2\$=\$\dfrac{1}{\sqrt2}\$)\$\mu_{2}\$ cos \$r_{2}=\mu_{3}\$ sin \$r_{3}\$ \$\Rightarrow\$ sin \$r_{3}\$ \$=\dfrac{\mu_{2}\sqrt{1-sin^{2}r_{2}}}{\sqrt{3}}\$\$\mu_{3}\$ cos \$r_{3}=1\$ = \$\dfrac{\mu_{2}^{2}-1}{\sqrt{3}}\$sin \$^{2} r_{3}\$ + cos \$^{2} r_{3}=1\$\$\Rightarrow \dfrac{\mu_{2}^{2}-1}{3}+\dfrac{1}{3}=1 \Rightarrow {\mu_{2}}^{2}=3\Rightarrow \mu_{2}=\sqrt{3}\$Thus we get \$x=3.\$

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 )\$

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Ray Optics And Optical Instruments

Q: Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

Angle of incident at the boundary = 60By snells law\$\Rightarrow \sqrt{3}\sin 60^{\circ}=1.5 \sin r\$\$\Rightarrow r=90^{\circ}\$Thus the angle of deviation as seen in the figure is given as \$90^o-60^o=30^o\$

Q: Two transparent media of refractive indices \$\mu_{1}\$ and \$\mu_{3}\$ have a solid lens shaped transparent material of refractive index \$\mu_{2}\$ between them as shown in figures in List 2. A ray traversing these media is also shown in the figures. In List 1 different relationships between \$\mu_{1},\ \mu_{2}\$ and \$\mu_{3}\$ are given. Match them to the ray diagram shown in List 2.

In the above equation, as sine is an increasing function in \$ 0 \$ to \$ \frac{\pi}{2} \$, if \$ \mu_{2} > \mu_{1}, i>r \$ and if \$ \mu_{2} < \mu_{1}, i<r \$.Thus light rays going from rarer medium to denser medium will bend towards the normal and vice versa.It can be seen from the image how the light rays bend when they refract at a curved surface. (Line along the radius is the normal here as shown in the figure)If the ray goes undeviated, then the refractive indexes are same.So in 1, \$ \mu_{1} \mu_{2} \$ and \$ \mu_{2} > \mu_{3} \$So in 3, \$ \mu_{1} \mu_{2} \$ and \$ \mu_{2} > \mu_{3} \$So in 5, \$ \mu_{1} > \mu_{2} \$ and \$ \mu_{2} = \mu_{3} \$

Q: A thin convex lens made of glass of \$\mu =\$ 1.5 has refracting surfaces of radii of curvature 10 cm each. The left space of lens contains air and the right space is filled with water of refractive index \$\dfrac{4}{3}\$. A parallel beam of light is incident on it. The position of the image is :

Data: \${ \mu }_{ a }=1,{ \mu }_{ g }=1.5,{ \mu }_{ w }=1.33,{ R }_{ 1 }=10cm,{ R }_{ 2 }=-10cm,u=-\infty \$For left surface,\$\dfrac{\mu_g}{v_1}-\dfrac{\mu_a}{u} = \dfrac{\mu_g-\mu_a}{R_1}\$For right surface,\$\dfrac{\mu_w}{v}-\dfrac{\mu_g}{v_1} = \dfrac{\mu_w-\mu_g}{R_2}\$From above two,\$ \dfrac { { \mu }_{ w } }{ v } -\dfrac { { \mu }_{ a } }{ u } =\dfrac { { \mu }_{ g }-{ \mu }_{ a } }{ { R }_{ 1 } } +\dfrac { { \mu }_{ w }-{ \mu }_{ g } }{ { R }_{ 2 } } \$or, \$ \dfrac { 1.33 }{ v } =\dfrac { 1.5-1 }{ 10 } -\dfrac { 1.33-1 }{ 10 } \$\$v=\dfrac { 1.33\times 10 }{ 0.66 } =20\ cm\$

Q: A ray \$OP\$ monochromatic light is incident on the face \$AB\$ of prism \$ABCD\$ near vertex \$B\$ at an incident angle of \$60^{o}\$ (see figure). If the refractive index of the material of the prism is \$\sqrt{3}\$, which of the following is (are) correct?

Using snell's law at surface AB, CD and AD\$\displaystyle sin^{-1}\frac{1}{\sqrt{3}} < sin^{-1} \frac{1}{\sqrt{2}}\$Net deviation is \$90^{o}\$

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: A ray of light is incident on a prism \$\mathrm{A}BC\$ of refractive index \$\sqrt{3}\$ as shown in figure.(a) Find the angle of incidence for which the deviation of light ray by the prism \$\mathrm{A}BC\$ is minimum.(b) By what angle the second prism must be rotated, so that the final ray suffer net minimum deviation.

Minimum deviation\$i_{1}=i_{2}=i\$ and \$r_{1}=r_{2}=r\$and \$r_{1}+r_{2}=A\$\$r=A/2=30^{\circ}\$Using Snell's law,\$\dfrac{sini_{1}}{sinr_{1}}=\mu\$\$sini_{1}=\sqrt{3}sin30^{\circ}\$\$sini_{1}=\dfrac{\sqrt{3}}{2}\$\$i_{1}=60^{\circ}\$When the prism is rotated anticlockwise by \$60^{\circ}\$ the combined prism will become a slab as shown in the figure 2. In this case the net deviation by the prism will be zero as the deviation by the two prism will be equal and opposite.

Physics
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Easy Qs Med Qs Hard Qs

Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

12 0^{\circ}

9 0^{\circ}

6 0^{\circ}

4 5^{\circ}

3 0^{\circ}

Hard

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Two transparent media of refractive indices $μ_{1}$ and $μ_{3}$ have a solid lens shaped transparent material of refractive index $μ_{2}$ between them as shown in figures in List 2. A ray traversing these media is also shown in the figures. In List 1 different relationships between $μ_{1}, μ_{2}$ and $μ_{3}$ are given. Match them to the ray diagram shown in List 2.

List 1

μ_{1} < μ_{2}

μ_{1} > μ_{2}

μ_{2} = μ_{3}

μ_{2} > μ_{3}

List 2

Hard

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A thin convex lens made of glass of $μ =$ 1.5 has refracting surfaces of radii of curvature 10 cm each. The left space of lens contains air and the right space is filled with water of refractive index $\frac{4}{3}$ . A parallel beam of light is incident on it. The position of the image is :

16 c m

20 c m

24 c m

10 c m

Hard

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A ray of light from a liquid $(μ = 3)$ is incident on a system of two right angled prism of refractive indices $3 a n d 2$ as shown in the figure. The light suffers zero net deviation when it emerges into air from surface CD. If the angle of incidence (in degrees) is $5 n$ . Find n ?

Hard

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Assertion

The moon appears purely white during the day but after sunset its colour is yellowish hue.

Reason

The scattered blue component of white light is mixed with yellow, the colour becomes white.

Assertion is True. Reason is also True and is the correct explanation of the Assertion.

Assertion is True. Reason is also True but is not the correct explanation of Assertion.

Assertion is True and Reason is False

Assertion is False and Reason is True

Hard

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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 $3$ , which of the following is (are) correct?

The ray gets totally internally reflected at face CD

The ray comes out through face AD

The angle between the incident ray and the emergent ray is 90

^{o}

The angle between the incident ray and the emergent ray is 120

^{o}

Hard

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The refracting angle of a prism $6 0^{o}$ .The refractive index of the material of the prism is $\frac{7}{3}$ . The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

3 0^{0}

4 5^{o}

4 0^{o}

5 0^{o}

Hard

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A ray of light enters at grazing angle of incidence into an assembly of three isosceles right-angled prisms having refractive indices $μ_{1} = 2, μ_{2} = x$ and $μ_{3} = 3$ . If finally emergent light ray also emerges at grazing angle then calculate x :

Hard

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Consider an optical system placed in water $(μ = 4 / 3)$ PQRS is a hollow glass slab filled with air $(μ = 1)$ This slab is kept on the principle axis of a lens in such a way that surface PS makes an angle $θ$ with the axis of the lens. $(0 < θ < 18 0^{\circ})$ , then

for transmission of light takes place through slab

θ_{m i n} = \frac{π}{2}

-

sin

^{- 1} (0.75)

for transmission of light through slab

θ_{m a x} = \frac{π}{2} +

sin

^{- 1} (0.75)

for

x = 50

cm final image is at 3m distance from O

for

x = 100

cm final image is at 3m distance from O

Hard

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A ray of light is incident on a prism $A B C$ of refractive index $3$ as shown in figure.
(a) Find the angle of incidence for which the deviation of light ray by the prism $A B C$ is minimum.
(b) By what angle the second prism must be rotated, so that the final ray suffer net minimum deviation.

3 0^{o}

6 0^{o}

6 0^{o}

3 0^{o}

6 0^{o}

6 0^{o}

None of these

Hard

View solution>

Ray Optics And Optical Instruments

Physics 2143 Views

Refractive indices of 2 different media with separating boundary at the diagonal of rectangular glass slab are shown. Total angle of deviation of the ray as shown in the figure, when it emerges in air is :

A thin convex lens made of glass of μ= 1.5 has refracting surfaces of radii of curvature 10 cm each. The left space of lens contains air and the right space is filled with water of refractive index 34​. A parallel beam of light is incident on it. The position of the image is :

A ray of light from a liquid (μ=3​) is incident on a system of two right angled prism of refractive indices 3​and2​ as shown in the figure. The light suffers zero net deviation when it emerges into air from surface CD. If the angle of incidence (in degrees) is 5n. Find n ?

The moon appears purely white during the day but after sunset its colour is yellowish hue.

The scattered blue component of white light is mixed with yellow, the colour becomes white.

A ray OP monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of 60o (see figure). If the refractive index of the material of the prism is 3​, which of the following is (are) correct?

The refracting angle of a prism 60o.The refractive index of the material of the prism is 37​​. The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

A ray of light enters at grazing angle of incidence into an assembly of three isosceles right-angled prisms having refractive indices μ1​=2​,μ2​=x​ and μ3​=3​ . If finally emergent light ray also emerges at grazing angle then calculate x :

Consider an optical system placed in water (μ=4/3) PQRS is a hollow glass slab filled with air (μ=1) This slab is kept on the principle axis of a lens in such a way that surface PS makes an angle θ with the axis of the lens. (0<θ<180∘), then

A ray of light is incident on a prism ABC of refractive index 3​ as shown in figure. (a) Find the angle of incidence for which the deviation of light ray by the prism ABC is minimum. (b) By what angle the second prism must be rotated, so that the final ray suffer net minimum deviation.

Physics
2143 Views

A thin convex lens made of glass of $μ =$ 1.5 has refracting surfaces of radii of curvature 10 cm each. The left space of lens contains air and the right space is filled with water of refractive index $\frac{4}{3}$ . A parallel beam of light is incident on it. The position of the image is :

A ray of light from a liquid $(μ = 3)$ is incident on a system of two right angled prism of refractive indices $3 a n d 2$ as shown in the figure. The light suffers zero net deviation when it emerges into air from surface CD. If the angle of incidence (in degrees) is $5 n$ . Find n ?

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 $3$ , which of the following is (are) correct?

The refracting angle of a prism $6 0^{o}$ .The refractive index of the material of the prism is $\frac{7}{3}$ . The limiting angle of incidence of a ray that will be transmitted through the prism in this case will be :

A ray of light enters at grazing angle of incidence into an assembly of three isosceles right-angled prisms having refractive indices $μ_{1} = 2, μ_{2} = x$ and $μ_{3} = 3$ . If finally emergent light ray also emerges at grazing angle then calculate x :

Consider an optical system placed in water $(μ = 4 / 3)$ PQRS is a hollow glass slab filled with air $(μ = 1)$ This slab is kept on the principle axis of a lens in such a way that surface PS makes an angle $θ$ with the axis of the lens. $(0 < θ < 18 0^{\circ})$ , then

A ray of light is incident on a prism $A B C$ of refractive index $3$ as shown in figure.
(a) Find the angle of incidence for which the deviation of light ray by the prism $A B C$ is minimum.
(b) By what angle the second prism must be rotated, so that the final ray suffer net minimum deviation.