Ray Optics and Optical Instruments Test

Result

Ray Optics and Optical Instruments Test - 6

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Different objects at different distances are seen by the eye. The parameter that remains constant is

A
the focal length of the eye lens

B
the object distance from the eye lens

C
the radii of curvature of the eye lens

D
the image distance from the eye lens

Solution

The image formed by the eye lens is always on the retina and the image distance is fixed.
Question 2
4 / -1

An under-water swimmer cannot see very clearly even in absolutely clear water because of

A
absorption of light in water

B
scattering of light in water

C
reduction of speed of light in water

D
change in the focal length of eye lens

Solution

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.
Question 3
4 / -1

The nearer point of hypermetropic eye is 40 cm. The lens to be used for its correction should have the power?

A
+1.5 D

B
-1.5 D

C
+2.5 D

D
+0.5 D

Solution

Hypermetropia is corrected by using convex lens.
Focal lenght of lens used f = +(defected near point)
f = +d = +40 cm
Question 4
4 / -1

A microscope is focused on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?

A
4.5 cm downward

B
1 cm downward

C
2 cm upward

D
1 cm upward

Solution

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
Question 5
4 / -1

A compound microscope consists of an objective lens with focal length 1.0 cm and eye piece of focal length 2.0 cm and a tube length 20 cm the magnification will be

A
100

B
200

C
250

D
300

Solution

Magnification, of compound microscope

Here, L = 20 cm, D = 25cm (near point),
f₀ = 1cm and f_e = 2cm
Question 6
4 / -1

In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm. If an object is placed at 2 cm from objective and the final image is formed at 25 cm from eye lens, the distance between the two lenses is

A
6.00 cm

B
7.75 cm

C
9.25 cm

D
11.0 cm

Solution

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
Question 7
4 / -1

A person with normal near point 25 cm using a compound microscope with objective of focal length 8.0 mm and an eye piece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. The separation between two lenses and magnification respectively are

A
9.47 cm, 88

B
3.36 cm, 44

C
6.00 cm, 22

D
7.49 cm, 11

Solution

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,
Question 8
4 / -1

The final image in an astronomical telescope with respect to object is

A
virtual and erect

B
real and erect

C
real and inverted

D
virtual and inverted

Solution

The final image is inverted and virtual as can be seen from the figure above.
Question 9
4 / -1

In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objeective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is

A
L/I

B

C

D

Solution

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.
Question 10
4 / -1

The focal length of the lenses of an astronomical telescope are 50 cm and 5 cm. The length of the telescope when the image is formed at the least distance of distinct vision is

A
45 cm

B
55 cm

C
275/6 cm

D
325/6 cm

Solution

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
Question 11
4 / -1

A small telescope has an objective lens of focal length 144 cm and an eye piece of focal length 6.0 cm. What is the separation between the objective and the eye piece?

A
0.75 m

B
1.38 m

C
1.0 m

D
1.5 m

Solution

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
Question 12
4 / -1

An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2cm.

A
the magnification is 1000

B
the length of the telescope tube is 20.02 m

C
the image formed is inverted

D
all of these

Solution

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.
Question 13
4 / -1

A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

A
1000

B
1500

C
2000

D
3000

Solution

Here, f₀= 15 m = 15 x 10² cm, f_e = 1.0 cm
Question 14
4 / -1

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. The magnifying power of the telescope for viewing distant objects when the final image is formed at the least distance of distinct vision 25 cm will be

A
33.6

B
66.12

C
22.6

D
11.6

Solution

When the final image is formed at least distance of distinct vision d, magnifying power of the telescope is
Question 15
4 / -1

A reflecting telescope has a large mirror for its objective with radius of curvature equal to 80 cm. The magnifying power of this telescope if eye piece used has a focal length of 1.6 cm is

A
100

B
50

C
25

D
5

Solution

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