JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Geometrical Optics

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The focal length of a thin biconvex lens is 20cm. When an object is moved from a distance of 25cm in front of it to 50cm, the magnification of its image changes from

m_{25}

m_{50}

. The ratio

\frac{m _{25}}{m _{50}}

Answer: 6

Solution

Using the lens formula:

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

with

f = 20

cm.

For u = -25 cm:

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

\frac{1}{v} = \frac{1}{20} - \frac{1}{25} = \frac{5 - 4}{100} = \frac{1}{100}

v = 100

cm

Magnification:

m_{25} = \frac{v}{u} = \frac{100}{- 25} = - 4

For u = -50 cm:

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

\frac{1}{v} = \frac{1}{20} - \frac{1}{50} = \frac{5 - 2}{100} = \frac{3}{100}

v = \frac{100}{3}

cm

Magnification:

m_{50} = \frac{v}{u} = \frac{100/3}{- 50} = - \frac{2}{3}

Ratio:

\frac{m _{25}}{m _{50}} = \frac{- 4}{- 2/3} = 4 \times \frac{3}{2} = 6

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Geometrical Optics

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