CBSE Class 12 Physics 2024: question 19
Set in the CBSE Class 12 board exam of 2024, paper 55/5/1, section B, question 19, for 2 marks. Try it first; the hints and our step-by-step solution are below.
Q19 · 2 marks · Very short answerRay Optics and Optical Instruments
A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. Calculate its magnifying power in normal adjustment and the distance of the image formed by the objective.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/5/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
In normal adjustment the final image is at infinity.
Hint 2
Where does a lens form the image of a very distant object?
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
In normal adjustment, magnifying power
. - 1 mark
The object (a distant one) is effectively at infinity, .
Lens formula for the objective: , so .
Hence cm. The image is formed at the focus of the objective, 150 cm behind it.
Answer: Magnifying power = 30; image formed by the objective at 150 cm from it
Where the marks are
- Magnifying power m = f_o/f_e = 30 (1 mark)
- Image distance of the objective = 150 cm (1 mark)
Revise the chapter
- Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)
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