The distance between an object and a divergent lens is x-times the focal length of the lens. The magnification m produced by the lens will be

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x+1

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\frac{1}{x}

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\frac{1}{(x + 1)}

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Solution

The correct option is D $\frac{1}{(x + 1)}$
Divergent lens $\Rightarrow concave lens Focal length = - f u = - x f By using lens formula \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \frac{1}{- f} = \frac{1}{v} - (\frac{1}{- x f}) = \frac{1}{v} + \frac{1}{x f} \frac{1}{v} = - \frac{1}{f} - \frac{1}{x f} \frac{1}{v} = \frac{- (x + 1)}{x f} v = \frac{- x f}{(x + 1)} Next we can find magnification m = \frac{v}{u} = \frac{- \frac{x f}{(x + 1)}}{- \frac{x f}{1}} m = \frac{- x f}{(x + 1)} \times \frac{1}{- x f} m = \frac{1}{(x + 1)}$

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The distance between an object and a divergent lens is x-times the focal length of the lens. The magnification m produced by the lens will be

The correct option is D 1(x+1)Divergent lens ⇒ concave lensFocal length=−fu=−xfBy using lens formula1f=1v−1u1−f=1v−(1−xf)=1v+1xf1v=−1f−1xf1v=−(x+1)xfv=−xf(x+1)Next we can find magnificationm=vu=−xf(x+1)−xf1m=−xf(x+1)×1−xfm=1(x+1)