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A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of height 2 cm placed 30 cm from the lens?

Virtual, upright, height=0.5 cm

Real, inverted, height=4 cm

Real, inverted, height=1 cm

Virtual, upright, height=1 cm

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Explanation

In general we have assumed $μ = 1.5$

So, $f = 20 cm$

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{1}{20} = \frac{1}{v} + \frac{1}{30} \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} v = 60 cm$

$\frac{h_{i}}{h_{o}} = \frac{v}{u} h_{i} = \frac{1}{2} \times | h_{o} | h_{i} = 4 cm$

Here, image is real, inverted, magnified field and height of image is 4 cm.

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Which of the following is not due to total internal reflection?

Difference between apparent and real depth of a pond

Mirage on hot summer days

Brilliance of diamond

Working of optical fibre

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Explanation

Real and apparent depth are explains on the basis of refraction only. The concept of TIR is not involed here.

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A thin prism of angle $15 °$ made of glass of refractive index $μ_{1} = 1.5$ is combined with another prism of glass of refractive index $μ_{2} = 1.75 .$ The combination of the prism produces dispersion without deviation. The angle of the second prism should be

$7 °$

$10 °$

$12 °$

$5 °$

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Explanation

For without deviation

$\frac{A}{A'} = \frac{μ' - 1}{μ - 1} \frac{15 °}{A'} = \frac{1.75 - 1}{1.5 - 1} \frac{15}{A'} = \frac{0.75}{0.50} A' = \frac{0.50 \times 15}{0.75} = 10 °$

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A converging beam of rays is incident on a diverging lens. Having passed though the lens the rays intersect at a point 15cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5cm closer to the lens. The focal length of the lens is:

-10cm

20cm

-30cm

5cm

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Explanation

Given u=10cm,v=15cm

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{f} = \frac{1}{15} - \frac{1}{10}$

$\frac{1}{f} = \frac{10 - 15}{150}$

$\frac{1}{f} = \frac{- 5}{150}$

f=-30cm

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A ray of light traveling in a transparent medium of refractive index $μ$ falls, on a surface separating the medium from the air at an angle of incidence of 45 $°$ .For which of the following value of $μ$ the ray can undergo total internal reflection?

$μ = 1.33$

$μ = 1.40$

$μ = 1.50$

$μ = 1.25$

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Explanation

For total internal reflection

i>c

$\Rightarrow \sin i > \sin c$

$\Rightarrow \sin 45 ° > \frac{1}{μ}$

$\Rightarrow μ > \sqrt{2}$

$\Rightarrow μ > 1.4$

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A lens having focal length f and aperature of diameter d forms an image of intensity I. Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

f and $\frac{1}{4}$

$\frac{3 f}{4} a n d \frac{I}{2}$

f and $\frac{3 I}{4}$

$\frac{f}{2} a n d \frac{I}{2}$

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Explanation

Intensity, $I \propto A^{2}$

$\Rightarrow \frac{I_{2}}{I_{1}} = {[\frac{A_{2}}{A_{1}}]}^{2} = \frac{{πr}^{2} - \frac{{πr}^{2}}{4}}{{πr}^{2}} = \frac{3}{4}$

$\Rightarrow I_{2} = \frac{3}{4} I_{1}$ and focal length remains unchanged.

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The speed of light in media $M_{1}$ and $M_{2}$ is $1.5 \times 10^{8} m / s$ and $2.0 \times 10^{8} m / s$ respectively. A ray of light enters from medium $M_{1}$ to $M_{2}$ at an incidence angle i. If the ray suffers total internal reflection, the value of i is

equal to $\sin^{- 1} (\frac{2}{3})$

equal to or less than $\sin^{- 1} (\frac{3}{5})$

equal to or greater than $\sin^{- 1} (\frac{3}{4})$

less than $\sin^{- 1} (\frac{2}{3})$

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Explanation

In total internal reflection, the angle of incidence (i) must be greater than critical angle C

$∵ μ_{1} = 2 and μ_{2} = \frac{3}{2} 2 \sin i \geq \frac{3}{2} \sin 90 ° \sin i \geq \frac{3}{4} i \geq \sin^{- 1} (\frac{3}{4})$

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A ray of light is incident on a $60 °$ prism at the minimum deviation position. The angle of refraction at the first face (ie, incident face) of the prism is

$zero$

$30 °$

$45 °$

$60 °$

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Explanation

The refracting angle of prism

$A = r_{1} + r_{2}$

For minimum deviation

$r_{1} = r_{2} = r ∴ A = 2 r or r = \frac{A}{2} = \frac{60 °}{2} = 30 °$

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Two thin lenses of focal length $f_{1} a n d f_{2}$ are in contact and coaxial. The power of the combination is

$\sqrt{\frac{f_{1}}{f_{2}}}$

$\sqrt{\frac{f_{2}}{f_{1}}}$

$\frac{f_{1} - f_{2}}{f_{1} f_{2}}$

$\frac{f_{1} + f_{2}}{f_{1} f_{2}}$

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Explanation

If two thin lenses of focal lengths $f_{1}, f_{2}$ are placed in contact coaxially, then equivalent focal length of combination is

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{0}{f_{1} f_{2}} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

Power for the combination is

$P = \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{f_{1} + f_{2}}{f_{1} f_{2}}$

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A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is 1.39 $\times 10^{9} m$ and its mean distance from the earth is $1.5 \times 10^{11} m$ . What is the diameter of the sun's image on the paper?

$9.2 \times 10^{- 4} m$

$6.5 \times 10^{- 4} m$

$6.5 \times 10^{- 5} m$

$12.4 \times 10^{- 4} m$

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Explanation

From the relation

$\frac{I}{O} = \frac{v}{u}$

$H e r e, O = 1.39 \times 10^{9} m, v = 0.1 m, u = 1.5 \times 10^{11} m ∴ I = \frac{0.1}{1.5 \times 10^{11}} \times 1.39 \times 10^{9} = 9.2 \times 10^{- 4} m$

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