PhysicsCBSE Class 12 · Board exam 2023, set 55/5/1 · 5 marks · Long answer

CBSE Class 12 Physics 2023: question 33(b)

Set in the CBSE Class 12 board exam of 2023, paper 55/5/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerRay Optics and Optical Instruments

(i) Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.
(ii) The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 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, so the two focal points coincide between the lenses.

Hint 2

Magnifying power is the ratio of the angle subtended by the image to the angle subtended by the object.

Hint 3

Form two equations, one from $m$ and one from the lens separation, and solve them together.

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.

(i) Ray diagram (described). Parallel rays from a distant object enter the objective, which has a large focal length $f_{o}$ , at an angle $α$ to the axis. They form a real, inverted image of height $h$ at the second focal plane of the objective. This focal plane coincides with the first focal plane of the eyepiece, so the lens separation is $f_{o} + f_{e}$ . Rays leave the eyepiece parallel at an angle $β$ to the axis, so the final image is formed at infinity (normal adjustment). Arrows on the rays show the direction of light.
1½ marks
Angle subtended by the object. Magnifying power is defined as $m = \frac{β}{α}$ . For small angles, the object (or the image at the objective) subtends $α \approx tan α = \frac{h}{f _{o}}$ .
½ mark
Angle subtended by the final image. The final image subtends $β \approx tan β = \frac{h}{f _{e}}$ at the eye.
½ mark
Expression. $m = \frac{β}{α} = \frac{h / f _{e}}{h / f _{o}} = \frac{f _{o}}{f _{e}}$ .
½ mark
(ii) Using the magnifying power. In normal adjustment, $m = \frac{f _{o}}{f _{e}} = 2.9$ , so $f_{o} = 2.9 f_{e}$ .
½ mark
Using the separation. The tube length is $f_{o} + f_{e} = 150$ cm.
½ mark
Solving. $2.9 f_{e} + f_{e} = 150 \Rightarrow 3.9 f_{e} = 150 \Rightarrow f_{e} \approx 38.5$ cm.
½ mark
Focal length of the objective. $f_{o} = 150 - 38.5 = 111.5$ cm.
½ mark

Answer: (i) $m = f_{o} / f_{e}$ ; (ii) $f_{o} \approx 111.5$ cm, $f_{e} \approx 38.5$ cm

Where the marks are

Correct ray diagram of an astronomical telescope in normal adjustment, with the final image at infinity (1½ marks)
Uses $α = h / f_{o}$ and $β = h / f_{e}$ to derive $m = f_{o} / f_{e}$ (1½ marks)
Writes $f_{o} / f_{e} = 2.9$ and $f_{o} + f_{e} = 150$ cm (1 mark)
Obtains $f_{e} \approx 38.5$ cm and $f_{o} \approx 111.5$ cm (1 mark)

Revise the chapter

Chapter: Ray Optics and Optical Instruments (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Ray Optics and Optical Instruments, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 55/5/1: every question, by section.

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CBSE Class 12 Physics 2023: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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