(a) If the image formed by a lens is diminished in size and erect, for all positions of the object, what type of lens is it?

(b) Name the point on the lens through which a ray of light passes undeviated.

(c) An object is placed perpendicular to the principal axis of a convex lens of focal length $20\text{ cm}$ . The distance of the object from the lens is $30\text{ cm}$ . Find (i) the position (ii) the magnification and (iii) the nature of the image formed.

OR

(a) One-half of a convex lens is covered with a black paper. Will such a lens produce an image of the complete object? Support your answer with a ray diagram.

(b) An object $5\text{ cm}$ high is held $25\text{ cm}$ away from a converging lens of focal length $10\text{ cm}$ .

(i) Draw the ray diagram and

(ii) Calculate the position and size of the image formed.

(iii) What is the nature of the image?

Light — Reflection and Refraction

Image formation by a convex lens and use of lens formula

Official Answer

(OR option answered)

(a) Yes, a convex lens with one half covered with black paper will still produce an image of the complete object, but the image will be less bright (of lower intensity). The uncovered half of the lens still refracts light from every point of the object and forms a complete image, although with only half the number of rays contributing to it.

(b) Given:

Height of object, h_o = +5 cm
Object distance, u = −25 cm (New Cartesian sign convention)
Focal length of converging lens, f = +10 cm

Using the lens formula:

1/f = 1/v − 1/u

1/10 = 1/v − 1/(−25)

1/v = 1/10 − 1/25 = (5 − 2)/50 = 3/50

v = 50/3 ≈ +16.67 cm

Magnification:

m = hi / ho = v / u = (50/3) / (−25) = −2/3 ≈ −0.667

Image height:

hi = m $\times$ ho = (−2/3) $\times$ 5 = −10/3 ≈ −3.33 cm

(i) Ray diagram: a ray from the top of the object travels parallel to the principal axis and after refraction passes through the focus F $_{2}$ on the other side; a second ray passes undeviated through the optical centre O; the two refracted rays meet at A', forming a real, inverted, diminished image A'B' on the other side of the lens.

(ii) Position: v = +16.67 cm (on the other side of the lens); Size: |h_i| = 3.33 cm.

(iii) Nature: Real, inverted, and diminished (since the object is between f and 2f).

convex lenslens formulaNew Cartesian sign conventionreal, inverted, diminishedmagnificationv = +16.67 cmh_i = −3.33 cm

Marking Scheme

11 mark for correctly answering (a) with reasoning and a labelled ray diagram.
21 mark for stating the lens formula and the given values with proper sign convention in (b).
31 mark for correctly calculating the image distance v = +16.67 cm.
41 mark for correctly calculating the magnification and the image size h_i = −3.33 cm.
51 mark for stating the nature of the image (real, inverted, diminished) and the correct ray diagram in (b).

Hint

Apply the New Cartesian sign convention carefully. A convex (converging) lens has a positive focal length, and a real object is placed to the left, so its distance is negative. Use the lens formula and the magnification formula m = v/u.

Quick Oral Answer

A half-covered lens still forms a complete image of the object, but the image is dimmer. For a 5 cm object placed 25 cm from a convex lens of focal length 10 cm, the image is formed at 16.67 cm on the other side, is 3.33 cm tall, and is real, inverted and diminished.

Analysis & Explanation

(a) A lens forms an image because rays from each point of the object are refracted and converge to a corresponding image point. Covering half the lens only blocks some of the rays; the rays passing through the remaining half still form the entire image, just with reduced brightness. This is the same reason an image can be formed by a pinhole — the full geometry is preserved.

(b) For a thin convex lens, the lens formula relates object distance u, image distance v and focal length f. The signs follow the New Cartesian Sign Convention: distances measured in the direction of the incident light are positive, and opposite to it are negative. With u = −25 cm and f = +10 cm, we get v = +50/3 cm ≈ +16.67 cm. The positive v confirms a real image on the opposite side of the lens. Magnification m = v/u = −2/3 indicates an inverted image of size 2/3 times the object. Hence the image is real, inverted and diminished. This is consistent with the rule: for a convex lens, when the object lies between f and 2f, the image is real, inverted, magnified and beyond 2f; when the object lies beyond 2f, the image is real, inverted and diminished between f and 2f. Here the object is at 25 cm, which is beyond 2f (20 cm), so the image is diminished between f and 2f on the other side — exactly matching our calculation.

Common Mistakes

1Taking the object distance as +25 cm instead of −25 cm, which leads to a wrong sign for v.
2Using m = u/v instead of v/u.
3Saying that covering half the lens does not produce any image, instead of stating that a complete but dimmer image is formed.
4Forgetting to state that the image is diminished (not enlarged) because the object is beyond 2f.

Interesting Facts

This is the same principle used in cameras: the convex lens forms a real, inverted, diminished image on the film or sensor.

A simple pinhole camera also forms a complete image because each point of the object sends rays through the small opening; a half-covered lens works on a related principle.

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2011?

This question carries 5 marks in the CBSE Class 10 Science 2011 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Light — Reflection and Refraction" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Image formation by a convex lens and use of lens formula" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2011 paper?

This is a Long Answer question from Section A in the CBSE Class 10 Science 2011 paper.