Answer the following. (i) Discuss briefly diffraction of light from a single slit and draw the shape of the diffraction pattern. (ii) An object is placed between the pole and the focus of a concave mirror. Using mirror formula, prove mathematically that it produces a virtual and an enlarged image.

(i) Central bright maximum with minima at aθ=nλ and fainter secondary maxima; (ii) v=(xF)/(F-x)>0 ⇒ virtual; m=(F)/(F-x)>1 ⇒ enlarged.

PhysicsCBSE Class 12 · Board exam 2024, set 55/5/1 · 5 marks · Case-based

CBSE Class 12 Physics 2024: question 33(b)

Set in the CBSE Class 12 board exam of 2024, paper 55/5/1, section E, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Case-basedCompetency-basedWave Optics

Answer the following.

(i)
Discuss briefly diffraction of light from a single slit and draw the shape of the diffraction pattern.
0 marks
(ii)
An object is placed between the pole and the focus of a concave mirror. Using mirror formula, prove mathematically that it produces a virtual and an enlarged image.
0 marks

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

For the slit, treat every point of the slit opening as a source of wavelets and ask when they arrive in phase or cancel.

Hint 2

Minima occur when the path difference between the two edges of the slit is a whole number of wavelengths; sketch intensity against $θ$ .

Hint 3

For the mirror, put $u = - x$ and $f = - F$ with $x < F$ into $v = u f / (u - f)$ and look at the sign of $v$ , then find $m = - v / u$ .

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) Diffraction: When a beam of monochromatic light falls normally on a narrow single slit of width $a$ (comparable to $λ$ ), it bends round the edges of the slit into the region of geometrical shadow. This bending is called diffraction. Each point of the wavefront at the slit acts as a source of secondary wavelets (Huygens). These wavelets superpose on the screen to give a pattern of bright and dark bands.
1 mark
At the centre O all the wavelets arrive in phase, so the central maximum is bright, with maximum intensity. At angle $θ$ with $a sin θ = nλ$ ( $n = \pm 1, \pm 2, \dots$ ), the slit can be divided into pairs of halves whose contributions cancel, giving minima. Secondary maxima occur near $a sin θ = (n + \frac{1}{2}) λ$ . Their intensity falls rapidly with increasing order. The central maximum is twice as wide as the others, with angular width $2 λ / a$ .
1 mark
Shape of pattern: Draw an intensity ( $I$ ) versus $θ$ graph. It has a tall, broad central peak at $θ = 0$ , zeros at $θ = \pm λ / a, \pm 2 λ / a, \dots$ , and much smaller secondary peaks of decreasing height on either side, symmetric about the centre.
1 mark
(ii) Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow v = \frac{u f}{u - f}$ . By the Cartesian sign convention, for a concave mirror $u = - x$ and $f = - F$ , with $0 < x < F$ (object between P and F). Then $v = \frac{( - x ) ( - F )}{- x + F} = \frac{x F}{F - x}$ . Since $F > x$ , $v$ is positive, so the image is formed behind the mirror. Hence the image is virtual (and erect).
1 mark
Magnification: $m = - \frac{v}{u} = - \frac{x F / ( F - x )}{- x} = \frac{F}{F - x}$ . Since $0 < F - x < F$ , $m > 1$ (and positive). Hence the image is enlarged and erect.
1 mark

Answer: (i) Central bright maximum with minima at $a sin θ = nλ$ and fainter secondary maxima; (ii) $v = \frac{x F}{F - x} > 0$ ⇒ virtual; $m = \frac{F}{F - x} > 1$ ⇒ enlarged.

Where the marks are

Explains diffraction at a single slit as bending of light round the edges and superposition of secondary wavelets (1 mark)
States conditions/intensity distribution: bright central maximum, minima at $a sin θ = nλ$ , secondary maxima of decreasing intensity (1 mark)
Draws intensity versus angle pattern with broad central maximum and weaker secondary maxima (1 mark)
Uses mirror formula with sign convention to show $v$ positive, so image is virtual (1 mark)
Shows $m = F / (F - x) > 1$ , so image is enlarged (1 mark)

Revise the chapter

Chapter: Wave Optics (notes, formulas, where marks are lost and every past question)
Topic: Single Slit Diffraction

Practise this chapterBoard-style questions from Wave Optics, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 12 Physics 2024: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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