Question

An object of height 4cm is placed on the principal axis at a distance of 20cm in front of a convex lens of focal length 30cm. Find the distance of image from the lens, its height and nature

Answer 1

Given:

An object of height 4cm is placed on the principal axis at a distance of 20cm in front of a convex lens of focal length 30cm.

To find:

• Image distance
• Image Height
• Image characteristics.

Calculation:

Applying Len's Formula:

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

$\implies \dfrac{1}{30} = \dfrac{1}{v} - \dfrac{1}{( - 20)}$

$\implies \dfrac{1}{30} = \dfrac{1}{v} + \dfrac{1}{ 20}$

$\implies \dfrac{1}{v} = \dfrac{1}{30} - \dfrac{1}{20}$

$\implies \dfrac{1}{v} = \dfrac{2 - 3}{60}$

$\implies \dfrac{1}{v} = - \dfrac{1}{60}$

$\implies \: v = - 60 \: cm$

So, image distance is 60 cm.

$\implies \: \dfrac{h_{i} }{h_{o} } = \dfrac{v}{u}$

$\implies \: \dfrac{h_{i} }{h_{o} } = \dfrac{ - 60}{ - 20}$

$\implies \: \dfrac{h_{i} }{h_{o} } = 3$

$\implies \: h_{i} = 3 \times 4 = 12 \: cm$

So, image height is 12 cm.

Image characteristics:

• Virtual
• Erect
• Magnified

Hope It Helps.