A lens of power + $3 D$ and another of power $- 1.5 D$ is placed in contact. What would be the focal length of the combination?

1.5 m

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- 0.67 m

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0.67 m

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- 67.0 m

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Solution

The correct option is C $0.67 m$
We know that , $p = \frac{1}{f_{i n m e t e r}}$

and we have, $f_{1} = \frac{1}{3}, f_{2} = - \frac{1}{1.5}$

Now, $\Rightarrow \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = 3 + - 1.5$

$\Rightarrow \frac{1}{f} = 1.5 f = \frac{1}{1.5} = 0.67 m$

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A lens of power + 3D and another of power −1.5D is placed in contact. What would be the focal length of the combination?

The correct option is C 0.67mWe know that , p=1finmeterand we have, f1=13,f2=−11.5Now, ⇒1f=1f1+1f2=3+−1.5⇒1f=1.5f=11.5=0.67m

A lens of power + $3 D$ and another of power $- 1.5 D$ is placed in contact. What would be the focal length of the combination?

The correct option is C $0.67 m$
We know that , $p = \frac{1}{f_{i n m e t e r}}$

and we have, $f_{1} = \frac{1}{3}, f_{2} = - \frac{1}{1.5}$

Now, $\Rightarrow \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = 3 + - 1.5$

$\Rightarrow \frac{1}{f} = 1.5 f = \frac{1}{1.5} = 0.67 m$