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Home/Practice questions/Physics
PhysicsRay Optics

In one trial, the candle is placed 15 cm from the lens. The magnifica…

Read the following passage and answer the questions.

Aman sets up a thin convex glass lens of focal length 20 cm on an optical bench. He places a lighted candle on one side of the lens and a white screen on the other side. He moves the candle to several positions. Each time, he slides the screen until a sharp image of the flame appears on it. For some positions of the candle, no image can be caught on the screen at all. He then removes the screen and looks at the candle through the lens. In one trial, he places the lens in a glass tank filled with water. In another trial, he places a thin concave lens in contact with the convex lens and looks for the new image. His teacher asks him to note the image distance and the size of the image each time. She also asks him to write the power of each lens in dioptre.

In one trial, the candle is placed 15 cm from the lens. The magnification of the image seen by Aman is

Answer & explanation+

Correct answer: option 3

The correct answer is Option (3) → +4

With the candle closer than the focal length, the lens gives a virtual image on the same side as the candle. The lens formula gives the image distance, and the magnification of a lens is \( m=v/u \).

Given:

    \( u=-15\ \mathrm{cm} \)

    \( f=+20\ \mathrm{cm} \)

    \( \begin{aligned} \frac{1}{v} &= \frac{1}{f}+\frac{1}{u} \\ &= \frac{1}{20}-\frac{1}{15} \\ &= -\frac{1}{60} \end{aligned} \)

    \( v=-60\ \mathrm{cm} \)

    \( \begin{aligned} m &= \frac{v}{u} \\ &= \frac{-60}{-15} \\ &= +4 \end{aligned} \)

The positive sign shows an erect, virtual image. Hence, the magnification of the image is +4.

NCERT (2026-27):
Refraction by a lens – “… Applying the sign convention, BO = – u, DI = +v, we get”

    \( \displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \qquad (9.23) \)

 

Refraction by a lens – “… it is easily seen that for a lens”

    \( \displaystyle m=\frac{h^{\prime}}{h}=\frac{v}{u} \)\( \qquad (9.24) \)

“When we apply the sign convention, we see that, for erect (and virtual) image formed by a convex or concave lens, m is positive, while for an inverted (and real) image, m is negative.”

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