The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio m_25 \over m_50 is \underline\hspace2cm .

Geometrical Optics PYQ — JEE Advanced 2010 · Shift 1 · Q81 with Solution

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Correct answer: 6

(Using Cartesian sign convention: object in front of lens has negative object distance.)

For a thin lens,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

and magnification is

m = \frac{v}{u}

We will calculate magnification in both cases.

Here,

u_1 = -25\ \text{cm}, \qquad f = 20\ \text{cm}

Using lens formula:

\frac{1}{20} = \frac{1}{v_1} - \left(-\frac{1}{25}\right)

\frac{1}{20} = \frac{1}{v_1} + \frac{1}{25}

\frac{1}{v_1} = \frac{1}{20} - \frac{1}{25}

\frac{1}{v_1} = \frac{5-4}{100} = \frac{1}{100}

So,

v_1 = 100\ \text{cm}

Now magnification:

m_{25} = \frac{v_1}{u_1} = \frac{100}{-25} = -4

Here,

u_2 = -50\ \text{cm}

Using lens formula:

\frac{1}{20} = \frac{1}{v_2} - \left(-\frac{1}{50}\right)

\frac{1}{20} = \frac{1}{v_2} + \frac{1}{50}

\frac{1}{v_2} = \frac{1}{20} - \frac{1}{50}

\frac{1}{v_2} = \frac{5-2}{100} = \frac{3}{100}

Thus,

v_2 = \frac{100}{3}\ \text{cm}

Now magnification:

m_{50} = \frac{v_2}{u_2} = \frac{100/3}{-50} = -\frac{2}{3}

\frac{m_{25}}{m_{50}} = \frac{-4}{-2/3}

\frac{m_{25}}{m_{50}} = -4 \times \left(-\frac{3}{2}\right) = 6

\boxed{6}

Stored correct answer = $6$

Our derived answer also equals $6$ , so they agree.