Question

(a) Why does one prefer to view the image formed at infinity than that formed at near point in microscope/telescope?

(b) Consider lenses $L_1$, $L_2$ and $L_3$ as specified in the following table. Which of them will you select as objective and eyepiece for constructing best possible (i) telescope (ii) compound microscope? Give reason for your answer.

Lens	$L_1$	$L_2$	$L_3$
Power	6 D	3 D	10 D
Aperture	1 cm	8 cm	1 cm

Hint:

Image at infinity keeps the eye relaxed. A telescope needs a long-f, large-aperture objective; a microscope needs two short-f lenses.

Answer 1

Step 1: Understanding the Question:
Part (a) asks where the final image is placed in a microscope or telescope and why. Part (b) asks us to pick lenses from their power and aperture.

Step 2: Part (a), image at infinity versus near point:
When the final image is at infinity, the rays reach the eye as parallel beams. The eye then looks at a far object, so its ciliary muscles stay relaxed.
When the image is at the near point (25 cm), the eye lens must bulge to its maximum. The ciliary muscles are under full strain, and the eye gets tired after a short time.
So for long, comfortable observation, the final image is set at infinity. The magnification is a little smaller than for the near point, but the viewing is comfortable.

Step 3: Part (b), focal lengths of the lenses:
The focal length is \( f = \dfrac{1}{P} \). Taking power in dioptre, we get:
\[ f_1 = \dfrac{1}{6}\ \text{m} = 16.7\ \text{cm},\quad f_2 = \dfrac{1}{3}\ \text{m} = 33.3\ \text{cm},\quad f_3 = \dfrac{1}{10}\ \text{m} = 10\ \text{cm} \]

Step 4: Part (b)(i), astronomical telescope:
The magnifying power of a telescope is \( m = \dfrac{f_o}{f_e} \). So the objective needs a large focal length and the eyepiece a small one.
The objective must also have a large aperture. A large aperture collects more light from a faint, distant object and gives a brighter, sharper image.
Only \( L_2 \) has a large aperture (8 cm) and a long focal length (33.3 cm), so \( L_2 \) is the objective.
For the eyepiece, \( L_3 \) (10 cm) gives \( m = 33.3/10 = 3.3 \), while \( L_1 \) (16.7 cm) gives only \( m = 33.3/16.7 = 2 \). So \( L_3 \) is the eyepiece. A 1 cm aperture is enough for an eyepiece.

Step 5: Part (b)(ii), compound microscope:
The magnifying power is \( m \approx \dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right) \). Both lenses must have short focal lengths, so \( L_2 \) is useless here.
The object lies very close to the objective, so a small aperture is acceptable. The shortest focal length is the best for the objective, so \( L_3 \) (10 cm) is the objective and \( L_1 \) (16.7 cm) is the eyepiece.

Final Answer:
(a) The image at infinity keeps the eye muscles relaxed. (b) Telescope: objective \( L_2 \), eyepiece \( L_3 \). Microscope: objective \( L_3 \), eyepiece \( L_1 \). \[ \boxed{\text{Telescope: } L_2,\ L_3;\quad \text{Microscope: } L_3,\ L_1} \]