Question

A convex lens forms a real image 3 times magnified when placed at a distance of 9 cm from the lens. Calculate.

1. Focal length of convex lens.
2. Power of convex lens.

Answer 1

Magnification (m) = 3

Distance of the object from the lens (u) = −9 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]

\[ \text{Applying, } m = \frac{v}{u} \]

\[ \Rightarrow 3 = \frac{v}{9} \]

\[ \Rightarrow v = 3 \times 9 \]

\[\Rightarrow v = 27 \ \text{cm}\]

As the image is real, therefore v = +27 cm.

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

\[ \Rightarrow \frac{1}{27} - \frac{1}{-9} = \frac{1}{f} \]

\[ \Rightarrow \frac{1}{f} = \frac{1}{27} + \frac{1}{9} \]

\[ \Rightarrow \frac{1}{f} = \frac{1 + 3}{27} \]

\[ \Rightarrow \frac{1}{f} = \frac{4}{27} \]

\[\ f = \frac{27}{4} \]

\[ \therefore \ f = + \ 6.75 \ \text{cm} \]

Thus, the focal length of the convex lens is 6.75 cm.

\[ \text{(b) Applying, } \text{P} = \frac{100}{f \text{(in cm)}} \]

\[ \Rightarrow \text{P} = \frac{100}{6.75} \]

\[ \Rightarrow \text{P} = + \ 14.81 \ \text{D} \]

Thus, the power of the convex lens is +14.81 D.