PhysicsCBSE Class 12 · Board exam 2025, set 55/1/1 · 5 marks · Long answer

CBSE Class 12 Physics 2025: question 33(b)

Set in the CBSE Class 12 board exam of 2025, paper 55/1/1, 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 · Long answerRay Optics and Optical Instruments

(i) Define a wavefront. An incident plane wave falls on a convex lens and gets refracted through it. Draw a diagram to show the incident and refracted wavefront.
(ii) A beam of light coming from a distant source is refracted by a spherical glass ball (refractive index 1.5) of radius 15 cm. Draw the ray diagram and obtain the position of the final image formed.

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

Treat the ball as two separate refracting spherical surfaces and use $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ twice.

Hint 2

The image from the first surface becomes the object for the second surface. Remember the two surfaces are 30 cm apart (the diameter).

Hint 3

At the second surface, light goes from glass to air, so swap $n_{1}$ and $n_{2}$ . Also take care with the sign of $R$ , which is now concave towards the incoming light.

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) A wavefront is the locus (surface) of all points of a medium that oscillate in the same phase.
1 mark
Diagram: Draw a convex lens with incident plane wavefronts as parallel straight lines perpendicular to the principal axis, approaching the lens.
½ mark
After refraction, the emergent wavefronts are spherical and converge, curving towards the focus F on the other side. The central part is delayed more because it travels through the thicker glass.
½ mark
(ii) Ray diagram: Draw parallel rays from the distant source falling on the first surface of the sphere (centre C, radius 15 cm). They bend towards the normal and head towards a point $I_{1}$ , 45 cm from the first surface. Before reaching $I_{1}$ they meet the second surface, bend away from the normal, and converge at $I$ , 7.5 cm beyond the second surface.
1 mark
Refraction at the first surface (air to glass): $n_{1} = 1$ , $n_{2} = 1.5$ , $u = \infty$ , $R = + 15$ cm.
$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$
$\frac{1.5}{v} - 0 = \frac{0.5}{15}$
$v = 45$ cm
So $I_{1}$ is 45 cm from the first surface.
1 mark
Refraction at the second surface (glass to air): $I_{1}$ lies $45 - 30 = 15$ cm beyond the second surface and acts as a virtual object. So $u = + 15$ cm, $R = - 15$ cm, $n_{1} = 1.5$ , $n_{2} = 1$ .
$\frac{1}{v} - \frac{1.5}{15} = \frac{1 - 1.5}{- 15}$
$\frac{1}{v} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30}$
$v = 7.5$ cm
1 mark
The final real image is formed 7.5 cm beyond the second surface of the ball, on the side away from the source. This is 22.5 cm from the centre of the ball.

Answer: Final image is real, 7.5 cm from the far surface of the ball (22.5 cm from its centre).

Where the marks are

Defines a wavefront as the locus of points oscillating in the same phase (1 mark)
Draws incident plane and refracted converging spherical wavefronts for a convex lens (½ + ½ mark)
Draws the ray diagram for refraction through the glass sphere (1 mark)
Applies the refraction formula at the first surface to get $v = 45$ cm (1 mark)
Takes the virtual object 15 cm beyond the second surface and obtains $v = 7.5$ 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 paper2025 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2025: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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