MCQ
An object is placed first at infinity and then at $20 cm$ from the object side focal plane of the convex lens. The two images thus formed are $5 cm$ apart. The focal length of the lens is......$cm$
- A$5$
- ✓$10$
- C$15 $
- D$20$
✓
Answer
Correct option: B.
$10$
b
(b) For first case : $\frac{1}{f} = \frac{1}{v} - \frac{1}{\infty } \Rightarrow f = v$
For second case $\frac{1}{f} = \frac{1}{{(f + 5)}} - \frac{1}{{ - (f + 20)}} \Rightarrow f = 10\,\,cm$
Alternative sol. -${f^2} = {x_1}{x_2}$ ==> $f = 10cm.$
(b) For first case : $\frac{1}{f} = \frac{1}{v} - \frac{1}{\infty } \Rightarrow f = v$
For second case $\frac{1}{f} = \frac{1}{{(f + 5)}} - \frac{1}{{ - (f + 20)}} \Rightarrow f = 10\,\,cm$
Alternative sol. -${f^2} = {x_1}{x_2}$ ==> $f = 10cm.$
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