Question

A concave lens has a focal length of 20 cm. At what distance from the lens, a 5 cm tall object is placed so that it forms an image at 15 cm from the lens?

Answer 1

✳ Required Answer:

✏ GiveN:

• It is a concave lens
• Focal length given = 20 cm
• Height of object = 5 cm
• Height of image = 15 cm

✏ To FinD:

• Object distance from lens...?

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✳ How to Solve?

The question require multiple concept like Magnification, Sign convention of Concave lens, mirror formula. Here's our required formulae to be used in this question.

✒ The mirror formula:

This formula shows the relation between object distance, image distance and focal length.

$\large{ \boxed{ \rm{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }}}$

Where, f is the focal length of lens, u is the object distance and v is the image distance.

✒Sign convention

• The focal length of convex lens is negative.
• The focal length of concave lens is positive.

✏Refer to the attachment...

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✳ Solution:

By using sign convention,

• f = -20 cm
• h(object) = 5 cm
• h(image) = 15 cm (Erect image)
• v = -15 cm (Concave lens forms virtual image)
• And u is negative because it is placed on the left of lens/mirror

By using mirror formula,

$\large{ \rm{ \longrightarrow \: \frac{1}{v} - \frac{1}{u} = \frac{1}{f} }} \\ \\ \large{ \rm{ \longrightarrow \: \frac{1}{ - 15} - \frac{1}{u} = \frac{1}{ - 20} }} \\ \\ \large{ \rm{ \longrightarrow \: - \frac{1}{u} = - \frac{ 1}{20} + \frac{1}{15} }} \\ \\ \large{ \rm{ \longrightarrow \: - \frac{1}{u} = \frac{ - 3 + 4}{60} }} \\ \\ \large{ \rm{ \longrightarrow \: - \frac{1}{u} = \frac{1}{60} }} \\ \\ \large{ \rm{ \longrightarrow \: u = \boxed{ \rm{ \red{ - 60 \: cm}}}}}$

So, the object should be placed on the left of the lens, because image distance is negative.

$\large{ \therefore{ \underline{ \underline{ \purple{ \rm{Hence \: solved \: \dag}}}}}}$

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