Question

The image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image.

Answer 1

According to the question; 

Object distance (u) = -45cm; 

Image distance (v) = 90cm; 

By lens formula;

\(\frac1v-\frac1u=\frac1f\)

⇒ \(\frac1{90}-\frac1{-45}=\frac1f\)

⇒ \(\frac1{f}=\frac{1+2}{90}=\frac3{90}\)

⇒ \(\frac1{f}=\frac1{90}\)

⇒ f = 30 cm.

Since f = 30cm which is positive hence it is a convex lens of focal length 30cm.

Now;

Height of Flame h₁= 2cm;

Magnification = \(\frac{h_2}{h_1}=\frac{v}u\) 

Putting values of v and u 

Magnification = \(\frac{h_2}{2}-\frac{90}{45}\)

⇒ \(\frac{h_2}{2}= -2;\)

⇒ h₂ = -2 x 2 = -4

Height of image is 4 cm. 

Negative sign means image is inverted.