Question

Q1. An object is placed perpendicular to
the principal axis of concave lens of
focal length 30 cm. Find the position of
an image when the object is at a
distance 20 cm from the lens.​

Answer 1

$\bigstar \mathfrak{\underline{\underline{Given:-}}}$

• An object is placed perpendicular to  the principal axis of concave lens.
• Focal length = -30 cm.
• Object distance=-20cm.

$\bigstar \mathfrak{\underline{\underline{To\ find:-}}}$

• Position of image.

$\bigstar \mathfrak{\underline{\underline{Solution :-}}}$

We know :

$\leadsto \underline{\boxed{\sf \frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}$

Where,

=> f = focal length

=> v=image distance

=> u=object distance

❥ As it is a concave lens,the focal length &object distance should be negative.

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Substituiting the given values in the formula:

$:\implies \sf \frac{1}{v} -\frac{1}{(-20)} =\frac{1}{-30}$

$:\implies \sf \frac{1}{v} +\frac{1}{20} =\frac{1}{-30}$

$:\implies \sf \frac{1}{v} =\frac{1}{-30} -\frac{1}{20}$

$:\implies \sf \frac{1}{v} =\frac{20+30}{-600} \\\\:\implies \sf \frac{1}{v} =\frac{1}{-12} \\\\:\implies \sf \boxed{\boxed{\bold{v=-12cm.}}}$

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Know More:-

❥ Sign conventions:-

All the distances are measured from the pole of the mirror.

Distances measured in the direction of incident rays are taken as positive.

Distances measured opposite to the direction of incident rays are taken as negative.

Distances measured upward and perpendicular to the principal axis are taken as positive.

Distances measured downward and perpendicular to the principal axis are taken as negative.

❥ Linear Magnification : Ratio of height of image to height of object.

❥ Use of concave lens: To treat myopia,used in telescopes,flashlights,cameras,etc.

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HOPE IT HELPS !