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Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Ray Optics and Optical Instruments

9.21 (a) 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?

(b) 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.

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Payal Gupta | Contributor-Level 10

4 months ago

9.21 Focal length of the convex lens, $f_{1}$ = 30 cm

Focal length of the concave lens, $f_{2}$ = -20 cm

Distance between two lenses, d = 8.0 cm

When the parallel beam of light is incident on the convex lens first:

According to lens formula, we have:

$\frac{1}{v_{1}}$ - $\frac{1}{u_{1}}$ = $\frac{1}{f_{1}}$ , where u = object distance = $\infty$ and $v_{1}$ = Image distance

$\frac{1}{v_{1}} =$ $\frac{1}{f_{1}} + \frac{1}{u_{1}}$ = $\frac{1}{30}$ + $\frac{1}{\infty}$

$v_{1} = 30 c m$

The image will act as a virtual object for the concave lens. Applying lens formula to the concave lens, we have:

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

where $u_{2}$ = object distance = $v_{1} - d$ = 30 – 8 = 22 cm and

$v_{2}$ = image distance

$\frac{1}{v_{2}} = \frac{1}{f_{2}} +$ $\frac{1}{u_{2}}$ =&

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9.21 Focal length of the convex lens, $f_{1}$ = 30 cm

Focal length of the concave lens, $f_{2}$ = -20 cm

Distance between two lenses, d = 8.0 cm

When the parallel beam of light is incident on the convex lens first:

According to lens formula, we have:

$\frac{1}{v_{1}}$ - $\frac{1}{u_{1}}$ = $\frac{1}{f_{1}}$ , where u = object distance = $\infty$ and $v_{1}$ = Image distance

$\frac{1}{v_{1}} =$ $\frac{1}{f_{1}} + \frac{1}{u_{1}}$ = $\frac{1}{30}$ + $\frac{1}{\infty}$

$v_{1} = 30 c m$

The image will act as a virtual object for the concave lens. Applying lens formula to the concave lens, we have:

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

where $u_{2}$ = object distance = $v_{1} - d$ = 30 – 8 = 22 cm and

$v_{2}$ = image distance

$\frac{1}{v_{2}} = \frac{1}{f_{2}} +$ $\frac{1}{u_{2}}$ = $\frac{1}{- 20}$ + $\frac{1}{22}$

$v_{2} =$ - 220 cm

The parallel incident beam appears to diverge from a point that is 220 - $\frac{d}{2}$ = 216 cm from the centre of the combination of two lenses.

When the parallel beam of light is incident, from the left on the concave lens first:

According to the lens formula,

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

$\frac{1}{v_{2}} = \frac{1}{f_{2}} + \frac{1}{u_{2}}$ , where $u_{2}$ = Object distance = - $\infty$

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

$v_{2} =$ - 20 cm

The image will act as a real object for the convex lens.

Applying lens formula to the convex lens, we have

$\frac{1}{v_{1}}$ - $\frac{1}{u_{1}}$ = $\frac{1}{f_{1}}$ , where $u_{1}$ = object distance = - (20 + d) = - 28 cm and $v_{1}$ = Image distance

$\frac{1}{v_{1}} = \frac{1}{f_{1}}$ + $\frac{1}{u_{1}}$ = $\frac{1}{30}$ + $\frac{1}{- 28}$

$v_{1}$ = - 420 cm

Hence, the parallel incident beam appear to diverge from a point that is 420 – 4 = 416 cm from the left of the centre of the combination of the two lenses.

The answer does not depend on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

Height of the image, $h_{1}$ = 1.5 cm

Object distance from the side of the convex lens, $u_{1}$ = - 40 cm

$|u_{1}|$ = 40 cm

According to lens formula:

$\frac{1}{v_{1}}$ - $\frac{1}{u_{1}}$ = $\frac{1}{f_{1}}$ , where $v_{1}$ = Image distance

$\frac{1}{v_{1}}$ = $\frac{1}{f_{1}} + \frac{1}{u_{1}}$ , where $v_{1}$ is the image distance

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

$v_{1}$ = 120 cm

$M a g n i f i c a t i o n, m_{1} =$ $\frac{v_{1}}{|u_{1}|}$ = 3

Hence the magnification due to the convex lens is 3.

The image formed by the convex lens acts as an object for the concave lens.

According to 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

$\frac{1}{v_{2}} =$ $\frac{1}{f_{2}} + \frac{1}{u_{2}}$ = $\frac{1}{- 20}$ + $\frac{1}{112}$

$v_{2}$ = - 24.35 cm

Magnification, $m_{2}$ = $|\frac{v_{2}}{112}|$ = 0.217

The magnification of combination of two lenses given as m = $m_{1} \times m_{2}$ = 0.652

The magnification of the combination is also expressed as

$\frac{h_{2}}{h_{1}}$ = 0.652, $h_{2}$ = 0.652 $\times h_{1}$ = 0.652 $\times 1.5$ = 0.978 cm

Hence, the height of the image is 0.978 cm.

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