Telescope Numerical Class-12 Nootan ISC Physics Solution Ch-18 Optical Instruments. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Telescope Numerical Class-12 Nootan ISC Physics Solution Ch-18 Optical Instruments
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-18 | Optical Instruments |
| Topics | Numericals on Telescope |
| Academic Session | 2025-2026 |
Numericals on Telescope
Telescope Numerical Class-12 Nootan ISC Physics Solution Ch-18 Optical Instruments.
Que-12: On seeing with unaided eye, the visual angle of moon at the eye is 0.6°. The focal lengths of the objective and the eye-piece of a telescope are respectively 200 cm and 5 cm. What will be the visual angle on seeing moon through the telescope?
Ans- m = β/α = fo/fe
=> 200/5 = β/0.6
=> β = 24°
Que-13: The focal lengths of the objective and the eye-piece of an astronomical telescope are 2.5 m and 5 cm respectively. What is the magnifying power of the telescope when the image is formed at infinity?
Ans- m = -fo/fe
=> -250/5 = -50
Que-14: The focal lengths of the objective and the eye-piece of an astronomical telescope are 75 cm and 5 cm respectively. Calculate the magnifying power and the length of the telescope when the final image is formed at (i) least distance of distinct vision (25 cm), (ii) infinity. (ISC 2005)
Ans- (i) for at D
m = -fo/fe (1 + fe/D)
=> -75/5 (1 + 5/25) = -18
and L = fo + [(ve x fe) / (ve + fe)]
=> 75 + (25 x 5/30) = 79.17 cm
(ii) for infinity
m = -fo/fe
=> -75/5 = -15
and L = fo + fe
=> 75 + 5 = 80
Que-15: What will be the maximum length of a telescope formed by lenses of focal lengths 4.0 cm and 60 cm? What will be the magnifying power? Is it also maximum?
Ans- L = fo + fe
=> 60 + 4 = 64 cm
m = -fo/fe
=> -60/4 = -15
Que-16: An astronomical telescope having a magnifying power of 6 for relaxed eye consists of two thin lenses 35 cm apart. Find the focal lengths of the lenses.
Ans- fo/fe = 6
=> fo = 6 fe
again fo + fe = 35
=> 6 fe + fe = 35
=> fe = 35/7 = 5 cm
and fo = 35-5 = 30 cm
Que-17: The focal length of the objective of an astronomical telescope is 1.0 m. If the magnifying power of telescope is 20, then determine the focal length of the eye-piece and the length of the telescope for relaxed eye.
Ans- m = -fo/fe
=> 100/fe = 20
=> fe = 10/200 = 5 cm
L = fo + fe
=> 100 + 5 = 105 cm
or 1.05 m
Que-18: A telescope objective has a focal length 1.00 m. When the final image is formed at the least distance of distinct vision (25 cm), the distance between the lenses is 1.05 m. Calculate the focal length of the eye-piece and the magnifying power of the telescope.
Ans- for final image at least distance of distinct vision
L = fo + ue = fo + [ve.fe / ve+fe]
=> L = 100 + [(25 x fe) / (25 + fe)] = 105
=> 25.fe / 25+fe = 5
=> fe = 25/4 = 6.25 cm
again m = -fo/fe(1 + fe/D)
=> -100/6.25 (1 + 6.25/25)
=> -100/6.25 x 5/4 = -20
Que-19: The objective of a telescope focussed for the least distance of distinct vision forms the image 0.75 m behind it. The distance between the two lenses is 0.80 m. What is the magnifying power of the telescope? For seeing with relaxed eye, by how much distance will the eye-piece have to be moved back? What will be the magnifying power then?
Ans- L = fo + ue = fo + [ve.fe / ve+fe]
=> L = 75+ [(25 x fe) / (25 + fe)] = 80
=> 25.fe / 25+fe = 5
=> fe = 25/4 = 6.25 cm
now m = -fo/fe(1 + fe/D)
=> -75/6.25 (1 + 6.25/25)
=> -75/6.25 x 5/4 = -15
again L for relaxed eye 75 + 6.25 = 81.25
∴ distance between lenses should be increased by 81.25 – 80 = 1.25 cm
and m = -fo/fe = -75/6.25 = -12
Que-20: An astronomical telescope has an objective of focal length 200 cm and an eye-piece of focal length 10 cm. An observer adjusts the distance of the eye-piece from the objective to obtain the image of the sun on a screen 40 cm behind the eye-piece. The diameter of the sun’s image is 6.0 cm. Find the diameter of the sun. The average earth-sun distance is 1.5 x 10^11 m.
Ans- 2 / (1.5 x 10^11) = d/D {as objective makes image at focus}
=> d = 2D / (1.5 x 10^11)
again for eye piece
1/10 = 1/40 – 1/u
=> u = -40/3
now v/u = I
=> 40/(40/3) = 0.06/d
=> 40/(40/3) = 0.06 x 1.5 x 10^11 / 2D
=> D = 1.5 x 10^9 m
— : End of Telescope Numerical Class-12 Nootan ISC Physics Solution Ch-18 Optical Instruments :–
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