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

CBSE Class 12 Physics 2023: question 33(a)

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

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

(i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point. (3)
(ii) In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope. (2)

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

Treat the microscope as two stages: the objective forms a real image, and the eyepiece magnifies that image like a simple magnifier.

Hint 2

For the final image at the near point, the eyepiece magnification is $1 + D / f_{e}$ .

Hint 3

Use the lens formula to find $v_{o}$ , then use $m_{o} = v_{o} / u_{o}$ .

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). A small object AB stands on the principal axis just beyond the focus $F_{o}$ of the objective (short focal length $f_{o}$ ). Rays from B pass through the objective and form a real, inverted, magnified image A′B′. A′B′ lies just inside the focus $F_{e}$ of the eyepiece. Rays from B′ then pass through the eyepiece and diverge. When produced backwards they meet at B″, giving a virtual, highly magnified, inverted final image A″B″ at the near point, distance $D$ from the eyepiece. Arrows on the rays show light travelling from object to eye.
1½ marks
Magnification by the objective. Let $L$ be the tube length, i.e. the distance between the second focal point of the objective and the first focal point of the eyepiece. Take $h$ as the object height and $h^{'}$ as the height of the first image. From similar triangles, $tan β = \frac{h}{f _{o}} = \frac{h ^{'}}{L}$ , so $m_{o} = \frac{h ^{'}}{h} = \frac{L}{f _{o}}$ .
½ mark
Magnification by the eyepiece. The eyepiece acts as a simple microscope with the final image at the near point, so $m_{e} = 1 + \frac{D}{f _{e}}$ .
½ mark
Total magnification. $m = m_{o} m_{e} = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})$ . Equivalently, $m = \frac{v _{o}}{∣ u _{o} ∣} (1 + \frac{D}{f _{e}})$ .
½ mark
(ii) Image by the objective. Given $u_{o} = - 1.5$ cm and $f_{o} = 1.25$ cm. Using $\frac{1}{v _{o}} - \frac{1}{u _{o}} = \frac{1}{f _{o}}$ : $\frac{1}{v _{o}} = \frac{1}{1.25} - \frac{1}{1.5} = \frac{2}{15}$ , so $v_{o} = 7.5$ cm.
½ mark
Magnification of the objective. $∣ m_{o} ∣ = \frac{v _{o}}{∣ u _{o} ∣} = \frac{7.5}{1.5} = 5$ .
½ mark
Magnification of the eyepiece. $m_{e} = 1 + \frac{D}{f _{e}} = 1 + \frac{25}{5} = 6$ .
½ mark
Magnifying power. $m = m_{o} \times m_{e} = 5 \times 6 = 30$ . With sign convention this is $- 30$ , the negative sign showing that the final image is inverted.
½ mark

Answer: (i) $m = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})$ ; (ii) magnifying power = 30

Where the marks are

Correct ray diagram of a compound microscope with the final image at the near point and arrows on the rays (1½ marks)
Writes $m_{o} = L / f_{o}$ and $m_{e} = 1 + D / f_{e}$ , and obtains $m = \frac{L}{f _{o}} (1 + \frac{D}{f _{e}})$ (1½ marks)
Finds $v_{o} = 7.5$ cm and $m_{o} = 5$ (1 mark)
Finds $m_{e} = 6$ and total $m = 30$ (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(a)

Hints

Step-by-step solution

Revise the chapter

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