QQuestion
An object placed in front of a convex lens forms an image of the same size on a screen. Moving the object 12 cm closer to the lens results in the formation of a real image which is three times the size of the object. Calculate the focal length of the lens.
✓Answer
Initially, the real image is the same size, so magnification = −1 and u = −v.
From the lens formula:
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{2}{\mathrm{v}}Therefore, v = 2f.
After moving the object 12 cm closer, u′ = u + 12 and magnification = −3. Hence, v′ = 3v − 36.
For the final position:
\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v'}}-\frac{1}{\mathrm{u'}}=\frac{4}{\mathrm{v'}}Therefore, f = v′/4 = (3v − 36)/4.
Using v = 2f:
f = \frac{6\mathrm{f}-36}{4}
Hence, f = 18 cm.
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