If the distances of an object and its virtual image from the focus of a convex lens of focal length f are 1 cm each, then f is

4 c m

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(\sqrt{2} + 1) c m

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2 \sqrt{2} c m

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(2 + \sqrt{2}) c m

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Solution

The correct option is B $(\sqrt{2} + 1) c m$

Here, u = –(f – 1)
v = –(f + 1)
f = +f
Applying $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ , we have
$\frac{1}{- (f + 1)} + \frac{1}{(f - 1))} = \frac{1}{f}$
or $f^{2} - 2 f - 1 = 0$
This gives $f = (\sqrt{2} + 1) c m$ .

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If the distances of an object and its virtual image from the focus of a convex lens of focal length f are 1 cm each, then f is

The correct option is B (√2+1) cm Here, u = –(f – 1) v = –(f + 1) f = +f Applying 1v−1u=1f, we have 1−(f+1)+1(f−1))=1f or f2–2f–1=0 This gives f=(√2+1) cm.

The correct option is B $(\sqrt{2} + 1) c m$

Here, u = –(f – 1)
v = –(f + 1)
f = +f
Applying $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ , we have
$\frac{1}{- (f + 1)} + \frac{1}{(f - 1))} = \frac{1}{f}$
or $f^{2} - 2 f - 1 = 0$
This gives $f = (\sqrt{2} + 1) c m$ .