Question

When an object is at distances $ {{u}_{1}} $ and $ {{u}_{2}} $ from a lens a real image and a virtual image is formed respectively having the same magnification. The focal length of the lens is

• A. $ {{u}_{1}}+\frac{{{u}_{2}}}{2} $
• B. $ \frac{{{u}_{1}}-{{u}_{2}}}{2} $
• C. $ \frac{{{u}_{1}}+{{u}_{2}}}{2} $
• D. $ {{u}_{1}}+{{u}_{2}} $

Answer 1

Answer: (C)

: For a convex lens when real image is formed, $ {{u}_{1}} $ is negative, $ {{v}_{1}} $ is positive, $ f $ is positive. Lens formula is $ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} $ $ \therefore $ For real image, $ \frac{1}{{{v}_{1}}}+\frac{1}{{{u}_{1}}}=\frac{1}{f} $ $ \therefore $ $ \frac{{{u}_{1}}}{{{v}_{1}}}+\frac{{{u}_{1}}}{{{u}_{1}}}=\frac{{{u}_{1}}}{f} $ $ \therefore $ $ \frac{{{u}_{1}}}{{{v}_{1}}}=\left( \frac{{{u}_{1}}}{f}-1 \right) $ ?.(i) When virtual image is formed, $ {{u}_{2}} $ is negative, $ {{v}_{2}} $ is also negative, $ f $ is positive $ \therefore $ For virtual image, $ \frac{1}{-{{v}_{2}}}+\frac{1}{{{u}_{2}}}=\frac{1}{f} $ Or $ \frac{{{u}_{2}}}{-{{v}_{2}}}+\frac{{{u}_{2}}}{{{u}_{2}}}=\frac{{{u}_{2}}}{f} $ or $ \frac{{{u}_{2}}}{{{v}_{2}}}-1=-\frac{{{u}_{2}}}{f} $ Or $ \frac{{{u}_{2}}}{{{v}_{2}}}=1-\frac{{{u}_{2}}}{f} $ ...(ii) Since magnification is same in the two cases. $ \frac{{{v}_{2}}}{{{u}_{2}}}=\frac{{{v}_{2}}}{{{u}_{1}}} $ or $ \frac{{{u}_{2}}}{{{v}_{2}}}=\frac{{{u}_{2}}}{{{v}_{1}}} $ .Substitute for these from (i) and (ii). Or $ \left( 1-\frac{{{u}_{2}}}{f} \right)=\left( \frac{{{u}_{1}}}{f}-1 \right) $ or $ 1+1=\frac{{{u}_{1}}}{f}+\frac{{{u}_{2}}}{f} $ Or $ 2=\frac{({{u}_{1}}+{{u}_{2}})}{f} $ or $ f=\frac{{{u}_{1}}+{{u}_{2}}}{2} $