Question

A concave lens forms 4 times diminished and virtual image when an object is placed at a distance of 80 cm. Calculate (a) the position of the image (b) the focal length of the lens.

Answer 1

Magnification (m) = `1/4` [image is 4 times diminished]

Distance of the object from the lens (u) = −80 cm [u is always −ive]

Distance of the image from the lens (v) = ? [to be calculated]

Focal length of the lens (f) = ? [to be calculated]

\[ \text{(a) Applying } m = \frac{v}{u} \]

\[ \Rightarrow \frac{1}{4} = \frac{v}{80} \]

\[ \Rightarrow v = \frac{80}{4} \]

\[ \Rightarrow v = 20 \ \text{cm} \]

As the image is virtual, therefore v = −20 cm.

Thus, the image is formed at a distance of 20 cm on the same side as the object.

\[ \text{(b) Applying } \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]

\[ \Rightarrow \frac{1}{-20} - \frac{1}{-80} = \frac{1}{f} \]

\[ \Rightarrow \frac{1}{f} = -\frac{1}{20} + \frac{1}{80} \]

\[ \Rightarrow \frac{1}{f} = \frac{-4 + 1}{80} \]

\[ \Rightarrow \frac{1}{f} = -\frac{3}{80} \]

\[ \Rightarrow f = -\frac{80}{3} \]

\[\Rightarrow f = - \ 26.67 \ \text{cm}\]

Thus, the focal length of the lens is −26.67 cm.