Question

The position of final image formed by the given lens combination from the third lens will be at a distance of (f₁ = + 10 cm, f₂ = – 10 cm and f₃ = + 30 cm).(a) 15 cm(b) infinity(c) 45 cm(d) 30 cm

Answer 1

I think 30cm is the answer

Answer 2

Answer:

The image distance of final image is formed at 30 cm.

(d) is correct option.

Explanation:

Given that,

Focal length of first lens = 10 cm

Focal length of second lens = -10 cm

Focal length of first lens = 30 cm

Object distance from first lens = 30 cm

We need to calculate the image distance for first lens

Using lens formula

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

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

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{-30}$

$\dfrac{1}{v}=\dfrac{1}{15}$

$v=15\ cm$

We need to calculate the image distance for second lens

Using lens formula

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

Here, u = 5-15 = -10 cm

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{-10}+\dfrac{1}{10}$

$\dfrac{1}{v}=\dfrac{1}{0}$

$v=\infty$

We need to calculate the image distance for third lens

Using lens formula

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

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{30}+\dfrac{1}{\infty}$

$\dfrac{1}{v}=\dfrac{1}{30}$

$v=30\ cm$

The final image is formed 30 cm to the right of third lens.

Hence, The image distance of final image is formed at 30 cm.