Answer the following questions

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Answer the following questions –

CBSE | Class 12 | NCERT | Physics | Ray Optics and Optical Instruments

Answer the following questions –

(i) Determine the ‘effective focal length’ of the combination of the two lenses in
Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does
the answer depend on which side of the combination a beam of parallel light is incident?
Is the notion of effective focal length of this system useful at all?

(ii) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement
(a) above. The distance between the object and the convex lens is 40 cm. Determine
the magnification produced by the two-lens system, and the size of the image.

Answer:

We are given –
Focal length of the given convex lens is $f_{1}$ = 30 cm
Focal length of the given concave lens is $f_{2}$ = −20 cm
Distance between the two lenses is d = 8.0 cm

(i) When the parallel beam of light is incident on the convex lens first, we have-
According to the lens formula, we can write:

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
Where,
$u_{1}$ is the Object distance
$v_{1}$ is the Image distance

$\frac{1}{v_{1}}=\frac{1}{30}-\frac{1}{\infty}=\frac{1}{30}$

Therefore, we have –

$v_{1}=30cm$

The image acts as a virtual object for a concave lens.
According to the lens formula for the concave lens, we can write –

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

Where,
$u_{2}$ is the Object distance
= (30 − d)

= 30 – 8

= 22 cm
$v_{2}$ is the Image distance

$\frac{1}{v_{2}}=\frac{1}{22}-\frac{1}{20}=\frac{10-11}{220}=\frac{-1}{220}$

Therefore, we have –

$v_{2}=-220cm$

From (420-4) 416cm, the parallel incident beam appears to diverge. The deviation occurs to the left of the centre of the two lenses’ combination. The answer is determined by which side of the combination has a parallel incident beam of light.
When the parallel beam of light is incident, on the concave lens, from the left first, we have –
According to the lens formula, we can write:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$

$\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}$
Where,
$u_{2}$ is the Object distance = −∞

$v_{2}$ is the Image distance

$\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{-\infty}=-\frac{1}{20}$

Therefore, we have –

$v_{2}=-20cm$

For a convex lens, the image acts as a real object.
Using the lens formula for the convex lens, we have –

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
Where,
$u_{1}$ is the Object distance
= −(20 + d)

= −(20 + 8)

= −28 cm
$v_{1}$ is the Image distance

$\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-28}=\frac{14-15}{420}=\frac{-1}{420}$

Therefore, we have –

$v_{2}=-420cm$

The parallel incident beam appears to diverge from (420-4) 416cm. The diversion happens from the left of the centre of the combination of the two lenses. The answer is dependent on the side of the combination where the incident beam of light is parallel.

(ii) Height of the image is $h_{1}$ = 1.5 cm
Object distance from the side of the given convex lens is $u_{1}$ = -40 cm
$|u_{1}|=40cm$

According to the lens formula:

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$
Where,

$v_{1}$ is the Image distance

$\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-40}=\frac{4-3}{120}=\frac{1}{120}$

Therefore, we have –

$v_{1}=120cm$

Magnification, m then becomes –

= $\frac{v_{1}}{|u_{1}|}=\frac{120}{40}=3$

Therefore, the magnification of the convex lens is 3.
The image of the convex lens is an object for the concave lens. Therefore, according to the lens formula:

$\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}$
Where,
$u_{2}$ is the Object distance
= +(120 − 8) = 112 cm.

$v_{2}$ is the Image distance

$\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{112}=\frac{-112+20}{2240}=\frac{-92}{2240}$

Therefore, we have –

$v_{2}=\frac{-2240}{92}$

Magnification then becomes –

m’= $|\frac{v_{2}}{u_{2}}|=\frac{2240}{92}\times \frac{1}{112}=\frac{20}{92}$

Therefore, the magnification due to the concave lens is $\frac{20}{92}$ .
Then the magnification produced by the combination of the two lenses is determined as: