Ray Optics And Optical Instruments

Q: A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12\ cm below the surface, the radius of this circle in cm is _____.

Since, the world is seen in the form of a circle, it implies, that the rays after refraction at the interface of the two media, must be going parallel to the surface. Applying Snell's law at the interface of the two media, we have, \mu_1\ Sin\ i = \mu_2\ Sin\ r Given, \mu_1 = \mu_{water} = \dfrac{4}{3} We know, \mu_2 = \mu_{air} = 1 From the figure, i = \theta and r = 90^{\circ} \Rightarrow \dfrac{4}{3}\ Sin \theta = 1 Sin\ \theta = \dfrac{3}{4} From the figure, Sin\ \theta = \dfrac{r}{\sqrt{r^2+h^2}} Given, h = 12\ cm \dfrac{r}{\sqrt{r^2+h^2}} = \dfrac{3}{4} Squaring both sides, we have \dfrac{r^2}{r^2+12^2} = \dfrac{9}{16} \dfrac{r^2 + 144}{r^2} = \dfrac{16}{9} \dfrac{144}{r^2} = \dfrac{16}{9} - 1 \Rightarrow \dfrac{144}{r^2} = \dfrac{7}{9} r^2 = \dfrac{144 \times 9}{7} r = \dfrac{36}{\sqrt{7}}\ cm Hence, the correct answer is OPTION D.

Q: A layer of oil of refractive index n_o floats on water of refractive index n_w . A beam of light is directed upward from a source in the water. Show that the critical angle for a total internal reflection is independent of n_o , even though, for angles close to C (critical angle), the total internal reflection always takes place at the oil-air surface.

Let the incidence angle at water-oil interface be i , and the refraction angle be r , which also acts as the incidence angle at oil-air interface.Then, at the water-oil interface by Snell's law: \dfrac{\sin r}{\sin i} = \dfrac{n_w}{n_o} \Rightarrow \sin r = \dfrac{n_w}{n_o}\sin i Again, for Total internal reflection at oil-air interface, by Snell's law: \dfrac{\sin 90^{\circ}}{\sin r}=\dfrac{n_o}{1} (since, n_{air} \approx 1 ) \Rightarrow \sin r = \dfrac{1}{n_o} \Rightarrow \dfrac{n_w}{n_o}\sin i = \dfrac{1}{n_o} \Rightarrow \sin i = \dfrac{1}{n_w} \therefore i = \sin^{-1}\left(\dfrac{1}{n_w}\right)=C Thus, the critical angle C for total internal reflection is independent of n_o .

Q: An optical fibre has a cylindrical cross-section of diameter d and index of refraction n bent sharply. What is the smallest radius of curvature at a short bent section for which the total internal reflection will be assured for light initially travelling parallel to the axis of the fibre?

Total internal reflection takes place for all angles greater than a critical angle. Thus, for the smallest radius of curvature, the angle made at the right end of the fibre (i) should be equal to the critical angle, for total internal reflection. Thus, \dfrac{\sin90^{\circ}}{\sin i} = \dfrac{n_1}{n_2} But, n_2 = 1 , and n_1 = n \Rightarrow \sin i = \dfrac{1}{n} \Rightarrow \dfrac{r-d}{r} = \dfrac{1}{n} \Rightarrow n(r-d) = r \Rightarrow nr - r = nd \Rightarrow r(n-1) = nd \therefore r = \dfrac{nd}{n-1}

Q: O is a point object kept on the principal axis of a concave mirror M of radius of curvature 20\ cm . P is a prism of angle 1.8^{\circ} . Light falling on the prism (at small angle of incidence) gets refracted through the prism and then falls on the mirror. Refractive index of the prism is 3/2 . Find the distance between the images formed by the concave mirror due to this light.

Given: O is a point object kept on the principal axis of a concave mirror M of radius of curvature 20 cm. P is a prism of angle 1.8^o . Light falling on the prism (at small angle of incidence) gets refracted through the prism and then falls on the mirror. The Refractive index of the prism is \dfrac 32 . To find the distance between the images formed by the concave mirror due to this light.Solution: For thin prism \delta=(\mu-1)\alpha =(1.5-1)1.8^\circ=(0.5)\times 1.8\times \dfrac \pi{180} rad=\dfrac\pi {200}rad Therefore distance, OO_1=10\times \dfrac \pi{200}=\dfrac \pi{20}cm Similarly, distance OO_2=\dfrac\pi{20}cm Therefore, O_1O_2=\dfrac\pi {10}cm For the mirror, O_1 and O_2 are two point objects, at a distance of 30 cm from mirror.Now, applying the mirror formula \dfrac 1v+\dfrac 1{-30}=\dfrac 1{-10}\implies v=-15cm Magnification is -\dfrac vu=-\dfrac {(-15)}{(-30)}=-\dfrac 12 (-ve sign indicates inverted image)Therefore, if O_1' and O_2' are the images formed, then distance between them, O_1'O_2'=\dfrac 12O_1O_2=\dfrac 12\times \dfrac \pi{10}=\dfrac \pi{20}cm

Q: The magnifying power of a compound microscope is 20 and the distance between its two lenses is 30 cm when the final image is at the near point of the eye. If the focal length of eyepiece is 6.25cm , the focal length of objective is :

M = 20 L = 30\ cm v_e = D= -25 \ cm f_e = 6.25 f_o = ? Using lens formula for eyepiece: \dfrac {1}{v_e} - \dfrac {1}{u_e} = \dfrac {1}{f_e} \dfrac {1}{-25} - \dfrac {1}{u_e} = \dfrac {1}{6.25} \dfrac {1}{-25} + \dfrac {1}{-6.25} = \dfrac {1}{u_e} u_e=5 \ cm From diagram length between lenses is: L = v_o + u_e 30 = v_o + 5 \Rightarrow v_o = 25 \ cm In compound microscope the magnification is given by: M = \dfrac {v_o}{u_o} (1 + \dfrac {D}{f_e}) 20 = \dfrac {25}{u_o} (1 + \dfrac {25}{6.25}) \Rightarrow u_o = \dfrac {25}{4} Using lens formula for objective lens: \dfrac {1}{v_o} - \dfrac {1}{u_o} = \dfrac {1}{f_o} \dfrac {1}{25} - \dfrac {4}{-25} = \dfrac {1}{f_o} \Rightarrow f_o = 5\ cm

Q: A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is \theta , then \theta close to.

The tower PQ in figure (19.5) subtends an angle \alpha on the objective. As u_o is very large, the first image P Q' is formed in the focal plane of the objective. tan \alpha=\alpha=\dfrac{P'Q'}{f\circ{}} ..(i)The final image P"Q" subtends an angle \beta on the eyepiece (and hence on the eye). We have from the triangle P Q'E tan \beta=\beta= \dfrac{P'Q'}{EP'} ...(ii)the telescope is set for normal adjustment so that the final image is formed at infinity, the first image P'Q' must be in the focal plane of the eyepiece.Then EP = {f_e} . Thus, equation (ii) becomes tan \beta=\beta= \dfrac{f\circ{}}{f_e} ...(iii)dividing equation (iii) by (i) \dfrac{\beta}{\alpha}=\dfrac{f\circ{}}{{f_e}} ...(iv)from equation (i) The \alpha =\dfrac{P'Q'}{f\circ{}} P'Q'= m h_1 ...(v) where m is magnification of objective lens m=\dfrac{v}{u} where v is position of first image P'Q' and u is position of tower v= f\circ{}=150 \;cm\;, u=1000\;m substituting value of m in equation (v) m P'Q'= \dfrac{.15}{1000} 50m=.0075 m \alpha =\dfrac{.0075}{.15}=.05 value of \beta can be obtained from equation (v) tan \beta =\dfrac{.15}{.05}\times \alpha=30\times.05=1.5 rad{} \beta = 56.3

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Difficult Questions

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Ray Optics And Optical Instruments

- Are you well prepared? Practice hard questions to be more confident about the chapter

The apparent depth of an object through layers of multiple media depends upon the refractive indices of the media and their thickness. Figure out the situation with a real world problem.

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is

4 / 3

and the fish is

12 c m

below the surface, the radius of this circle in cm is _____.

36 / 5

45

9

36 / 7

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A beam of light rays refracts when it passes through different mediums having different refractive indices.

Refraction, critical angle and total internal reflection come into picture. Let's see the problem

A layer of oil of refractive index

n_{o}

floats on water of refractive index

n_{w}

. A beam of light is directed upward from a source in the water. Show that the critical angle for a total internal reflection is independent of

n_{o}

, even though, for angles close to C (critical angle), the total internal reflection always takes place at the oil-air surface.

View solution

Total internal reflection occurs when the light rays travel from a more dense medium into a less dense medium (i.e. glass to air) and also the angle of incidence must be greater than the critical angle. Let's solve a question and learn how it works.

An optical fibre has a cylindrical cross-section of diameter

d

and index of refraction

n

bent sharply. What is the smallest radius of curvature at a short bent section for which the total internal reflection will be assured for light initially travelling parallel to the axis of the fibre?

View solution

When light rays pass through combination of prisms and mirrors, it get reflect and refract simultaneously. Multiple images are formed in this case. Let's see how it works.

O

is a point object kept on the principal axis of a concave mirror

M

of radius of curvature

20 c m

P

is a prism of angle

1 . 8^{\circ}

. Light falling on the prism (at small angle of incidence) gets refracted through the prism and then falls on the mirror. Refractive index of the prism is

3 / 2

. Find the distance between the images formed by the concave mirror due to this light.

View solution

When we place an object in front of a compound microscope a linear, angular and total magnification produced by a compound microscope. Try to understand this with a problem.

The magnifying power of a compound microscope is

20

and the distance between its two lenses is

30 c m

when the final image is at the near point of the eye. If the focal length of eyepiece is

6.25 c m

, the focal length of objective is :

2.5 c m

3.5 c m

4.5 c m

5.0 c m

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When we see an object with the help of telescope its magnified and resolved image is seen by us. It is completely depends upon the resolving power of the telescope. Let's see a problem and learn how it works.

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is

θ

, then

θ

close to.

6 0^{o}

1^{o}

3 0^{o}

1 5^{o}

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