Question

A 5cm tall object is placed on the principal axis of. convex lens of focal length 50cm at a distance of 40cm feom it . Find the nature , position and size of the image​

Answer 1

Points before solving:

• draw the ray diagram as per the question, which will give general idea of image formed.
• Given:

1. object distance = u = -40 cm
2. focal length(convex lens) = +50 cm

$\implies\rm lens\: formula \to \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$\implies\rm magnification = m = \dfrac{v}{u} =\dfrac{h_i}{h_o}$

Solution:

Finding image distance(v)?

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

$\to\rm \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

$\to\rm \dfrac{1}{v} = \dfrac{1}{+50} + \dfrac{1}{-40}$

$\to\rm \dfrac{1}{v} = \dfrac{1}{50} - \dfrac{1}{40}$

$\to\rm \dfrac{1}{v} = \dfrac{40-50}{50\times 40} = \dfrac{-10}{2000}$

$\to\bf v = -200\: cm$

★ Note:

• Negative sign implies image is formed at the same side of object,i.e, image is virtual.

Size of image:

$\to\rm m =\dfrac{v}{u} = \dfrac{200}{40}$

$\to\bf m = + 5$

$\to \rm\dfrac{h_i}{h_o} = 5$

$\to \rm h_i = 5\times 5\: cm = 25 \: cm$

Height of image is 25 cm tall.

★Note:

• positive sign indicates that image is formed up direction from principle axis,i.e,erect.
• image formed is 5 times magnified than object.

Image formed is:

1. virtual, erect.
2. 200 cm from the lens in the negative principle axis
3. magnified / enlarged .

Answer 2

Explanation:

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$\large\bf{\underline{\red{VERIFIED✔}}}$

Points before solving:

draw the ray diagram as per the question, which will give general idea of image formed.

Given:

object distance = u = -40 cm

focal length(convex lens) = +50 cm

$\implies\rm lens\: formula \to \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$\implies\rm magnification = m = \dfrac{v}{u} =\dfrac{h_i}{h_o}$

Solution:

Finding image distance(v)?

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

$\to\rm \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

$\to\rm \dfrac{1}{v} = \dfrac{1}{+50} + \dfrac{1}{-40}$

$\to\rm \dfrac{1}{v} = \dfrac{1}{50} - \dfrac{1}{40}$

$\to\rm \dfrac{1}{v} = \dfrac{40-50}{50\times 40} = \dfrac{-10}{2000}$

$\to\bf v = -200\: cm$

★ Note:

Negative sign implies image is formed at the same side of object,i.e, image is virtual.

Size of image:

$\to\rm m =\dfrac{v}{u} = \dfrac{200}{40}$

$\to\bf m = + 5$

$\to \rm\dfrac{h_i}{h_o} = 5$

$\to \rm h_i = 5\times 5\: cm = 25 \: cm$

Height of image is 25 cm tall.

★Note:

positive sign indicates that image is formed up direction from principle axis,i.e,erect.

image formed is 5 times magnified than object.

Image formed is:

virtual, erect.

200 cm from the lens in the negative principle axis

magnified / enlarged .

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