Test: Optical Instruments (NCERT) - NEET MCQ

The image formed by the eye lens is always on the retina and the image distance is fixed.

The eye lens is surrounded by a different medium than air. This will change the focal length of the eye lens. The eye cannot accommodate all images as it would do in air.

Hypermetropia is corrected by using convex lens.
Focal lenght of lens used f = +(defected near point)
f = +d = +40 cm

In the later case microscope will be focussed for O'. So, it is required to be lifted by distance OO'.
OO' = real depth of O - apparent depth of O

Magnification, of compound microscope

Here, L = 20 cm, D = 25cm (near point),
f₀ = 1cm and f_e = 2cm

Here, f₀ = 1.5 cm, f_e = 6.25cm, u₀ = −2 cm
v_e = −25cm
For objective,


For eye piece,



u_e = −5cm

Distance between two lenses = |v₀| + |u_e|
=  6cm + 5cm = 11cm

Here,d = 25cm, f₀ = 8.0 mm, f_e = 2.5 cm,u₀ = −9.0 mm = −0.9 cm

Therefore, separation between two lenses = u_e + v₀ = 2.27 + 7.2 = 9.47 cm 
Magnifying power, 

The final image is inverted and virtual as can be seen from the figure above.

Let f_o and f_e be the focal lengths of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length) = f_o + f_e. Treating the line on the objective as the object, and the eyepiece as the lens, u = -(f_o + f_e) and f = f_e.
1/v - 1/-(f_o + f_e) = 1/f_e
or 1/v = 1/f_e - 1/f_o + f_e = f_o/(f_o + f_e) f_e
or v = (f_o + f_e)f_e / f_o
Magnification =|v/u| = f_e/f_o = image size/object size = l/L:
∴ f_o/f_e = L/l = magnification of telescope in normal adjustment.

Here, f₀ = 50 cm, f_e = 5cm, D = 25cm
The length of the telescope when the image is formed at the least distance of distinct vision is



= 325/6 cm

The separation between the objective and the eye piece = Length of the telescope tube f = f_o + f_e
Here, f₀ = 144cm = 1.44m,
f_e = 6.0 cm = 0.06m
∴ = 1.44 + 0.06 = 1.5m

Here, f₀ = 20 m and f_e = 2 cm = 0.02m
In normal adjustment,
Length of telescope tube, L = f₀ + f_e = 20 + 0.02 = 20.02m
and magnification, m = f₀ / f_e = 20/0.02 = 1000
The image formed is inverted with respect to the object.

Here, f₀ = 15 m = 15 x 10² cm,  f_e = 1.0 cm

 

When the final image is formed at least distance of distinct vision d, magnifying power of the telescope is

The focal length of objective mirror
f₀ = R/2 = 80/2 = 40 cm
and focal length of eye piece = 1.6 cm
∴ magnifying power, m = f₀/f_e = 40/1.6 = 25