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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364534 Assertion :
A reflecting type telescope is preferred in astronomy.
Reason :
Weight of the mirrors are less as compared to the lenses.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364535 The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the telescope is $1.1 m$ . The magnification when the image is formed at half distance of the distinct vision ( $D = 25 c m$ ) is

1 6

2 14

3 18

4 16

Explanation:

$m = \frac{f_{0}}{f_{e}}; L = f_{0} + f_{e}$ $\Rightarrow f_{e} = \frac{L}{m + 1}$
$\Rightarrow f_{e} = 10 c m$
Magnifying power of a telescope is
$m = \frac{- f_{0}}{f_{e}} [1 + \frac{f_{e}}{| v_{e} |}]$
Where $| v_{e} |$ is the position of the final image
$= 10 [1 + \frac{10}{12.5}]$ $= \frac{225}{12.5} = 18$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364536 A telescope has an objective of focal length $50 c m$ and an eyepiece of focal length $5 c m$ . The least distance of distinct vision is 25. The telescope is focussed for distinct vision on a scale $200 c m$ away. The separation between the objective and the eyepiece is

1

75 c m

2

60 c m

3

71 c m

4

74 c m

Explanation:

Given, $f_{o} = 50 c m, f_{e} = 5 c m,$ $d = 25 c m, u_{o} = 200 c m, L = ?$
As, $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$
$∴ \frac{1}{v_{o}} = \frac{1}{f_{o}} + \frac{1}{u_{o}}$
$= \frac{1}{50} - \frac{1}{200} = \frac{4 - 1}{200} = \frac{3}{200}$
$v_{o} = \frac{200}{3} c m$
Linear magnification produced by
objective lens,
$m_{o} = \frac{v_{o}}{u_{o}} = - \frac{200 / 3}{200} = - \frac{1}{3}$
Now, $v_{e} = d = - 25 c m$
From $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$
$\Rightarrow \frac{- 1}{u_{e}} = \frac{1}{f_{e}} - \frac{1}{v_{e}}$
$= \frac{1}{5} + \frac{1}{25} = \frac{5 + 1}{25}$
$u_{e} = - \frac{25}{6} c m$
Distance between objective and
eye piece
$L = | v_{o} | + | u_{e} | = \frac{200}{3} + \frac{25}{6}$
$= \frac{400 + 25}{6} = \frac{425}{6}$
$= 70.8 c m \approx 71 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364537 While viewing a distant object with telescope a housefly sits on objective lens. Then

1 Housefly will be seen enlarged in image.

2 Housefly will be seen reduced in image.

3 Intensity of image will be decreased.

4 Intensity of image will be increased.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364534 Assertion :
A reflecting type telescope is preferred in astronomy.
Reason :
Weight of the mirrors are less as compared to the lenses.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364535 The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the telescope is $1.1 m$ . The magnification when the image is formed at half distance of the distinct vision ( $D = 25 c m$ ) is

1 6

2 14

3 18

4 16

Explanation:

$m = \frac{f_{0}}{f_{e}}; L = f_{0} + f_{e}$ $\Rightarrow f_{e} = \frac{L}{m + 1}$
$\Rightarrow f_{e} = 10 c m$
Magnifying power of a telescope is
$m = \frac{- f_{0}}{f_{e}} [1 + \frac{f_{e}}{| v_{e} |}]$
Where $| v_{e} |$ is the position of the final image
$= 10 [1 + \frac{10}{12.5}]$ $= \frac{225}{12.5} = 18$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364536 A telescope has an objective of focal length $50 c m$ and an eyepiece of focal length $5 c m$ . The least distance of distinct vision is 25. The telescope is focussed for distinct vision on a scale $200 c m$ away. The separation between the objective and the eyepiece is

1

75 c m

2

60 c m

3

71 c m

4

74 c m

Explanation:

Given, $f_{o} = 50 c m, f_{e} = 5 c m,$ $d = 25 c m, u_{o} = 200 c m, L = ?$
As, $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$
$∴ \frac{1}{v_{o}} = \frac{1}{f_{o}} + \frac{1}{u_{o}}$
$= \frac{1}{50} - \frac{1}{200} = \frac{4 - 1}{200} = \frac{3}{200}$
$v_{o} = \frac{200}{3} c m$
Linear magnification produced by
objective lens,
$m_{o} = \frac{v_{o}}{u_{o}} = - \frac{200 / 3}{200} = - \frac{1}{3}$
Now, $v_{e} = d = - 25 c m$
From $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$
$\Rightarrow \frac{- 1}{u_{e}} = \frac{1}{f_{e}} - \frac{1}{v_{e}}$
$= \frac{1}{5} + \frac{1}{25} = \frac{5 + 1}{25}$
$u_{e} = - \frac{25}{6} c m$
Distance between objective and
eye piece
$L = | v_{o} | + | u_{e} | = \frac{200}{3} + \frac{25}{6}$
$= \frac{400 + 25}{6} = \frac{425}{6}$
$= 70.8 c m \approx 71 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364537 While viewing a distant object with telescope a housefly sits on objective lens. Then

1 Housefly will be seen enlarged in image.

2 Housefly will be seen reduced in image.

3 Intensity of image will be decreased.

4 Intensity of image will be increased.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364534 Assertion :
A reflecting type telescope is preferred in astronomy.
Reason :
Weight of the mirrors are less as compared to the lenses.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364535 The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the telescope is $1.1 m$ . The magnification when the image is formed at half distance of the distinct vision ( $D = 25 c m$ ) is

1 6

2 14

3 18

4 16

Explanation:

$m = \frac{f_{0}}{f_{e}}; L = f_{0} + f_{e}$ $\Rightarrow f_{e} = \frac{L}{m + 1}$
$\Rightarrow f_{e} = 10 c m$
Magnifying power of a telescope is
$m = \frac{- f_{0}}{f_{e}} [1 + \frac{f_{e}}{| v_{e} |}]$
Where $| v_{e} |$ is the position of the final image
$= 10 [1 + \frac{10}{12.5}]$ $= \frac{225}{12.5} = 18$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364536 A telescope has an objective of focal length $50 c m$ and an eyepiece of focal length $5 c m$ . The least distance of distinct vision is 25. The telescope is focussed for distinct vision on a scale $200 c m$ away. The separation between the objective and the eyepiece is

1

75 c m

2

60 c m

3

71 c m

4

74 c m

Explanation:

Given, $f_{o} = 50 c m, f_{e} = 5 c m,$ $d = 25 c m, u_{o} = 200 c m, L = ?$
As, $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$
$∴ \frac{1}{v_{o}} = \frac{1}{f_{o}} + \frac{1}{u_{o}}$
$= \frac{1}{50} - \frac{1}{200} = \frac{4 - 1}{200} = \frac{3}{200}$
$v_{o} = \frac{200}{3} c m$
Linear magnification produced by
objective lens,
$m_{o} = \frac{v_{o}}{u_{o}} = - \frac{200 / 3}{200} = - \frac{1}{3}$
Now, $v_{e} = d = - 25 c m$
From $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$
$\Rightarrow \frac{- 1}{u_{e}} = \frac{1}{f_{e}} - \frac{1}{v_{e}}$
$= \frac{1}{5} + \frac{1}{25} = \frac{5 + 1}{25}$
$u_{e} = - \frac{25}{6} c m$
Distance between objective and
eye piece
$L = | v_{o} | + | u_{e} | = \frac{200}{3} + \frac{25}{6}$
$= \frac{400 + 25}{6} = \frac{425}{6}$
$= 70.8 c m \approx 71 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364537 While viewing a distant object with telescope a housefly sits on objective lens. Then

1 Housefly will be seen enlarged in image.

2 Housefly will be seen reduced in image.

3 Intensity of image will be decreased.

4 Intensity of image will be increased.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364534 Assertion :
A reflecting type telescope is preferred in astronomy.
Reason :
Weight of the mirrors are less as compared to the lenses.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364535 The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the telescope is $1.1 m$ . The magnification when the image is formed at half distance of the distinct vision ( $D = 25 c m$ ) is

1 6

2 14

3 18

4 16

Explanation:

$m = \frac{f_{0}}{f_{e}}; L = f_{0} + f_{e}$ $\Rightarrow f_{e} = \frac{L}{m + 1}$
$\Rightarrow f_{e} = 10 c m$
Magnifying power of a telescope is
$m = \frac{- f_{0}}{f_{e}} [1 + \frac{f_{e}}{| v_{e} |}]$
Where $| v_{e} |$ is the position of the final image
$= 10 [1 + \frac{10}{12.5}]$ $= \frac{225}{12.5} = 18$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364536 A telescope has an objective of focal length $50 c m$ and an eyepiece of focal length $5 c m$ . The least distance of distinct vision is 25. The telescope is focussed for distinct vision on a scale $200 c m$ away. The separation between the objective and the eyepiece is

1

75 c m

2

60 c m

3

71 c m

4

74 c m

Explanation:

Given, $f_{o} = 50 c m, f_{e} = 5 c m,$ $d = 25 c m, u_{o} = 200 c m, L = ?$
As, $\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$
$∴ \frac{1}{v_{o}} = \frac{1}{f_{o}} + \frac{1}{u_{o}}$
$= \frac{1}{50} - \frac{1}{200} = \frac{4 - 1}{200} = \frac{3}{200}$
$v_{o} = \frac{200}{3} c m$
Linear magnification produced by
objective lens,
$m_{o} = \frac{v_{o}}{u_{o}} = - \frac{200 / 3}{200} = - \frac{1}{3}$
Now, $v_{e} = d = - 25 c m$
From $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$
$\Rightarrow \frac{- 1}{u_{e}} = \frac{1}{f_{e}} - \frac{1}{v_{e}}$
$= \frac{1}{5} + \frac{1}{25} = \frac{5 + 1}{25}$
$u_{e} = - \frac{25}{6} c m$
Distance between objective and
eye piece
$L = | v_{o} | + | u_{e} | = \frac{200}{3} + \frac{25}{6}$
$= \frac{400 + 25}{6} = \frac{425}{6}$
$= 70.8 c m \approx 71 c m$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

364537 While viewing a distant object with telescope a housefly sits on objective lens. Then

1 Housefly will be seen enlarged in image.

2 Housefly will be seen reduced in image.

3 Intensity of image will be decreased.

4 Intensity of image will be increased.