एक प्रयोग के दौरान, \(15\; cm\) फोकस दूरी के एक उत्तल लेंस को एक प्रकाश बेंच पर एक उत्तल दर्पण के सामने \(5 \;cm\) दूरी पर समाक्षीय स्थिति में रखा गया है। यदि वस्तु को लेंस से \(20\; cm\) की दूरी पर रखा जाये तो वस्तु तथा उसका प्रतिबिम्ब संपाती हो जाते हैं। उत्तल दर्पण की फोकस दूरी \(\dots \; cm\) होगी

object is placed at a distance \((u)=-20\, \mathrm{cm}\)

By lens formula,

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

\(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)

\( = \frac{1}{{15}} - \frac{1}{{20}}\)

\(\frac{1}{v}=\frac{4-3}{60}\)

\({v}=60\, \mathrm{cm}\)

The image \(I\) gets formed at \(60\,\mathrm{cm}\) to the right of the lens and it wil be inverted.

The rays from the image \((I)\) formed further falls on the convex mirror forms another image This image should formed in such a way that it coincide with object at the same point due to reflection takes place by convex mirror.

Distance between lens and mirror will be \(\mathrm{d}\,=\) imagedistance \((\mathrm{v})\,-\) radius of curvature of convex mirror.

\(5=60-2 f\)

\(2 f=60-5\)

\(\mathrm{f}=\frac{55}{2}=27.5\, \mathrm{cm}\) (convex mirror)