Suppose Sita has kept a needle in front of a concave mirror of focal length f at a distance (f + X) and a real image of the needle is seen on a screen at a distance (f + y). Then the focal length f can be expressed as:

Correct option is B

The correct answer is (B)  f = $$\sqrt{xy}$$  

Given:

u = -(f + x)

v = -(f + y)  ( because concave mirror)

Formula used:

​$$\frac{1}{f}$$ = $$\frac{1}{u}$$ + $$\frac{1}{v}$$

Solution:

​$$\frac{1}{-f}$$ = $$\frac{1}{-(f + x)}$$ + $$\frac{1}{-(f + y)}$$

$$\frac{1}{f}$$ = $$\frac{1}{(f + x)}$$ + $$\frac{1}{(f + y)}$$

$$\frac{1}{f}$$ = $$\frac{(f + y) + (f + x)}{(f + x)(f + y)}$$

f (2f + y +x) = (f + y)(f + x)

2$$f^2$$ + (y + x)f = $$f^2$$ + (y + x)f + xy

2$$f^2$$ = $$f^2$$ + xy

$$f^2$$ = xy

f = $$\sqrt{xy}$$

Hence, the Focal Length is f = $$\sqrt{xy}$$​

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