A beam of a parallel rays is brought to a focus by a convex lens. If a thin concave lens of equal focal length is joined to the convex lens, the focus will be

Shifted to infinity

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Shifted to a small distance

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Remained undisturbed

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Shifted to the back side

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Solution

The correct option is A Shifted to infinity
Let the focal length of convex lens be $f_{1} = f$
If thin concave lens of same focal length then its focal length be $f_{2} = - f$
When these two lens are joined then
$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow f = \frac{f_{1} \times f_{2}}{f_{1} + f_{1}}$
By putting values we get
$f = \frac{f \times (- f)}{f + (- f)}$
$\Rightarrow f \to \infty$

Option A is correct.

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Q. A convex lens of focal length

1 m e t r e

and a concave lens of focal length

0.25 c m

are kept

0.75 m

apart. A parallel beam of light first passes through the convex lens, then through the concave lens and comes to a focus

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away from the concave lens.

(a) With the help of a labelled diagram explain how a convex lens converges a beam of parallel light rays. Mark the principal axis, optical centre, principal focus and focal length of the convex lens on the diagram.
(b) State whether the convex lens has a real focus or a virtual focus. (c) List some things that convex lens and concave mirror have in common.

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A beam of a parallel rays is brought to a focus by a convex lens. If a thin concave lens of equal focal length is joined to the convex lens, the focus will be

The correct option is A Shifted to infinityLet the focal length of convex lens be f1=fIf thin concave lens of same focal length then its focal length be f2=−fWhen these two lens are joined then 1f=1f1+1f2⇒f=f1×f2f1+f1By putting values we getf=f×(−f)f+(−f)⇒f→∞Option A is correct.