Question

A convex lens forms a real image of the object when the lens is at a distance of 1 m form the object. The size of the image is 3 times that of the object. The position of the object and the screen are kept fixed but the distance of lens from the object is doubled. The image is now one half of the size of the object, the focal length of the lens is

• A. 1.6 m
• B. 0.80 m
• C. 0.40 m
• D. 0.20 m.

Answer 1

The Correct Option is C

f = $ d/ (m_1 - m_2)$. Here d = 1 m, m, = 3 and $m_2$ = 1 / 2. Hence f = 40 cm

Answer 2

Given:
Initial object distance from the lens, d = 1m.
Initial magnification, m₁ = 3 (image size is 3 times the object size).
After doubling the object distance, d = 2m.
Magnification in the new situation, m₂ = \(\frac{1}{2}\) (image size is half the object size).

We need to find the focal length f of the lens:
\(f = \frac{d}{m_1 - m_2}\)

Calculate f using the initial conditions:
\(f = \frac{1}{3 - \frac{1}{2}}\)

\(f = \frac{1}{3 - 0.5}\)

\(f = \frac{1}{2.5}\)

\(f = 0.4 \text{ m}\)

So, the correct option is (C): 0.40m