Question

A lens placed between a candle and a screen forms a real triply magnified image of the candle on the screen. When the lens is moved away from the candle by 0.8 m without changing the positions of candle and screen, a real image one-third the size of the candle is formed on the screen. Determine the focal length of the lens (in cm).​

Answer 1

Answer:

30cm

Explanation:

As per given information

X = u2 − u1

= 0.8 m  

D = u1 + v1

We are given that magnification is 3 times;

m1 = v1/u1 =3

v1 = 3u1

When moved away, it gets 1/3rd

v2/u2 = 1/3

v2 = u2 / 3

x = v1 − u1

= 3u1−u1 = 0.8

u1 = 0.4

v1 = 3×0.4

   =1.2

D =1.2+ 0.4 =1.6  

f = D^2 – x^2 / 4D

= [(1.6)2−(0.8)2 ] / 4×1.6

= 0.3m

= 30cm

Answer 2

The focal length of the lens is 0.3 m

Given:

x = u₂ - u₁ = 0.8 m

Explanation:

The magnification is given by the formula:

m₁ = v₁/u₁ = 3 ⇒ v₁ = 3u₁

v₂/u₂ = 1/3 ⇒ v₂ = u₂/3

Now, x = v₁ - u₁

x = 3u₁ - u₁ = 2u₁ = 0.8

∴ u₁ = 0.4 m

⇒ v₁ = 3 × (0.4)

∴ v₁ = 1.2 m

The power of the lens is given as:

D = 1.2 + 0.4 = 1.6 m

The focal length is given as:

f = (D² - x²)/4D

On substituting the values, we get,

f = ((1.6)² - (0.8)²)/(4 × 0.8)

∴ f = 0.3 m