Test: Optical Instruments - JEE MCQ

The aperture of an astronomical telescope is defined as the diameter of the objective lens.
In telescopes since the stars are very far away from us and emit light intensities, we need a large aperture to increase the amount of light entering the telescope thereby increasing the resolution.

The resolving power of a telescope is given by RP∝2μsinθ​/122λ
Further, sinθ ∝ tanθ ∝ diameter of the objective.
So, the resolving power of a telescope depends on the diameter of the objective lens.

Long-sighted people who have lost their spectacles can still read a book by looking through a small (3-4 mm) hole in a sheet of paper because by doing so due to the diffraction at the hole the focal length of the eye lens is effectively decreased. Hence the correct option is C.
 

The final image of the moon should be formed at a distance of 100 cm. Using a concave lens of focal length 100 cm will enable him  to see the moon distinctly.

Focal length of lens =-100 cm

Time taken for 360° shift = 24 h
Time taken for 1° shift = 24/360 h = 4 min.
 

Area of magnification, 
A₁​A₂=μ²
A₁/6​​=100
A_1​=600 cm²

Magnification in microscope is  m=− lD/f₀​​f_e
where, l is length of tube and D is least distance for clear vision
f_0​ is focal length of objective
fe​ is the focal length of the eyepiece.
So, as l increases m increases.
Hence, the magnifying power of a microscope increases with increase in tube length.
 

f₀/f_e ​​=9,         ∴f₀​=9f_e​
Also f_0​+f_e​=20             (∵ final image is at infinity)
9f_e​+f_e​=20,f_e​=2cm,
∴f₀​=18cm

'u' should be the least distance of distinct vision.
U=-25cm
V=-50
(1/f)=(1/v)-(1/u)
=(-1/50) + (1/25)
=1/50
F=50cm
Hence a convex lens of focal length 50cm should be used.