PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 2 marks · Very short answer

CBSE Class 12 Physics 2025: question 18(a)

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, section B, question 18(a), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q18(a) · 2 marks · Very short answerWave Optics

In a diffraction experiment, the slit is illuminated by light of wavelength 600 nm. The first minimum of the pattern falls at $θ = 3 0^{\circ}$ . Calculate the width of the slit.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Which condition gives the angular positions of the dark fringes in single-slit diffraction?

Hint 2

For the first minimum use $a sin θ = λ$ , then put in $sin 3 0^{\circ} = 1/2$ .

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.

Condition for the $n^{t h}$ minimum in single-slit diffraction: $a sin θ = nλ$ . For the first minimum $n = 1$ : $a sin θ = λ$ .
1 mark
$a \times sin 3 0^{\circ} = 600 \times 1 0^{- 9}$ m, so $a \times \frac{1}{2} = 600 \times 1 0^{- 9}$ m.
½ mark
$a = 1200 \times 1 0^{- 9}$ m $= 1.2 \times 1 0^{- 6}$ m $(1.2 μ$ m $)$ .
½ mark

Answer: a = 1.2 × 10⁻⁶ m (1.2 μm)

Where the marks are

Writes the first-minimum condition $a sin θ = λ$ (1 mark)
Substitutes $sin 3 0^{\circ} = \frac{1}{2}$ and $λ = 600$ nm (½ mark)
Gets $a = 1.2 \times 1 0^{- 6}$ m (½ 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 paper2025 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2025: question 18(a)

Hints

Step-by-step solution

Revise the chapter

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