Question

An object is placed at the distance of 50cm from a concave lens of focal length 20 cm. Find the nature and position of the image.

Answer 1

Answer:

Focal length of the lens is 33.3 cm

Explanation:

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Answer 2

Given:

• Object distance = 50 cm
• Focal length = 20 cm
• Lens is concave

To find:

• Nature & Position of image?

Calculation:

• Focal length of concave lens will be negative.

Applying LEN'S FORMULA:

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\implies \dfrac{1}{ - 20} = \dfrac{1}{v} - \dfrac{1}{ - 50}$

$\implies - \dfrac{1}{ 20} = \dfrac{1}{v} + \dfrac{1}{ 50}$

$\implies \dfrac{1}{v} = - \dfrac{1}{ 50} - \dfrac{1}{20}$

$\implies \dfrac{1}{v} = \dfrac{ - 2 - 5}{ 100}$

$\implies \dfrac{1}{v} = \dfrac{ -7}{ 100}$

$\implies v = - 14.28 \: cm$

So, image is formed 14.28 cm in front of lens.

• Now , magnification will be :

$mag. = \dfrac{v}{u} = \dfrac{ - 14.28}{ - 50} < 1$

So, image is virtual, erect and diminished.