A convergent beam of light passes through a diverging lens of focal length 0-2 m and comes to focus 0-3 m behind the lens- The position of the point at which the beam would converge in the absence of the lens is

The correct option is A 0.12 m Here, f = 0.2 m, v = +0.3 mThe lens formula 1v 1u = 1f ⇒ 1u = 1v 1f = 10.3 + 10.2 = 0.50.006 ...

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Standard X

Physics

Lens Formula and Magnification

A convergent ...

Question

A convergent beam of light passes through a diverging lens of focal length $0 \cdot 2 m$ and comes to focus $0 \cdot 3 m$ behind the lens. The position of the point at which the beam would converge in the absence of the lens is

0.12 m

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0.6 m

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0.3 m

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0.15 m

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Solution

The correct option is A $0.12 m$
Here, f = -0.2 m, v = +0.3 m
The lens formula
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{0.3} + \frac{1}{0.2} = \frac{0.5}{0.006}$
$u = \frac{0.006}{0.5} = 0.12 m$

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A convergent beam of light passes through a diverging lens of focal length 0⋅2 m and comes to focus 0⋅3 m behind the lens. The position of the point at which the beam would converge in the absence of the lens is

The correct option is A 0.12 mHere, f = -0.2 m, v = +0.3 mThe lens formula1v−1u=1f⇒1u=1v−1f=10.3+10.2=0.50.006u=0.0060.5=0.12m

A convergent beam of light passes through a diverging lens of focal length $0 \cdot 2 m$ and comes to focus $0 \cdot 3 m$ behind the lens. The position of the point at which the beam would converge in the absence of the lens is

The correct option is A $0.12 m$
Here, f = -0.2 m, v = +0.3 m
The lens formula
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{0.3} + \frac{1}{0.2} = \frac{0.5}{0.006}$
$u = \frac{0.006}{0.5} = 0.12 m$