Ncert Solutions Physics Class 12th
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9.32 Figure 9.34 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
9.32 Figure 9.34 shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of the liquid?
9.32 Focal length of the convex lens, = 30 cm
The liquid acts as a mirror, focal length of the liquid =
Focal length of the system (convex lens + liquid), = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as
= &nb
Upvote 
9.31 Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.33. A current in the coil produces a deflection of 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
9.31 Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.33. A current in the coil produces a deflection of 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
9.31 Angle of deflection, = 3.5
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection, i.e. 2
The displacement (d) of the reflected spot of light on the screen is given as:
=
d = 1.5 tan 7 &nbs
Upvote 9.30 A Cassegrain telescope uses two mirrors as shown in Fig. 9.30. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
9.30 A Cassegrain telescope uses two mirrors as shown in Fig. 9.30. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
9.30 Distance between the objective mirror and the secondary mirror, d = 20 mm
Radius of curvature of objective mirror, = 220 mm
Hence focal length of the objective mirror, = = 110 mm
Radius of curvature of secondary mirror, = 140 mm
Hence focal length of t
Upvote 9.29 (a) For the telescope described in Exercise 9.28 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
9.29 (a) For the telescope described in Exercise 9.28 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
9.29 Focal length of the objective lens, = 140 cm
Focal length of the eyepiece, = 5 cm
Least distance of distinct vision, d = 25 cm
In normal adjustment, the separation between the objective lens and the eyepiece
=
Height of the tower
Distance of the tower (object) from t
Upvote 9.28 A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) The telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) The final image is formed at the least distance of distinct vision (25 cm)?
9.28 A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) The telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) The final image is formed at the least distance of distinct vision (25 cm)?
9.28 Focal length of the objective lens, = 140 cm
Focal length of the eyepiece, = 5 cm
Least distance of distinct vision, d = 25 cm
When the telescope is in normal adjustment, its magnifying power is given as:
= = = 28
When the final image is formed at
Upvote 9.27 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
9.27 An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
9.27 Focal length of the objective lens, = 1.25 cm
Focal length of the eyepiece, = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyep
Upvote 9.26 Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
9.26 Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
9.26 Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the object placed closer than the least distance of distinct vision (i.e. 25 cm). A closer object causes a larger angular size. A magnifyin
Upvote 9.25 What should be the distance between the object in Exercise 9.24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.23 to 9.25 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
9.25 What should be the distance between the object in Exercise 9.24 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
[Note: Exercises 9.23 to 9.25 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
9.25 Area of the virtual image of each square, A = 6.25
Area of each square, = 1
Hence the linear magnification of the object can be calculated as:
m = = = 2.5, but m = or v = 2.5u
Focal length of the magnifying glass, f = 10 cm. According to lens
Upvote 9.24 (a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
9.24 (a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
9.24 The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm)
Image distance, v = -d = -25 cm
Focal length, f = 10 cm
Object distance = u
According the lens formula, we have:
= -
= - = -
u = - 7.14 cm
Hence
Upvote 9.23 A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
9.23 A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
9.23 Area of each square, A = 1
Object distance, u = - 9 cm
Focal length of the converging lens, f = 10 cm
For image distance v, the lens formula can be written as
= -
= -
=
v = -90 cm
Magnification, m = = = 10
Therefore th
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