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Q1 A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? Ans: Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = −27 cm
Radius of curvature of the concave mirror, R = −36 cm
Focal length of the concave mirror, f = R/2 = -18 cm
Image distance = v
The image distance can be obtained using the mirror formula:
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.
Q2 A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. Ans: Height of the needle, h1 = 4.5 cm
Object distance, u = −12 cm
Focal length of the convex mirror, f = 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
Hence, magnification of the image, m = h2/h1 = 2.5/4.5 = 0.56
The height of the image is 2.5 cm.
The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.
Q10 What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses. Ans: Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = −20 cm
Focal length of the system of lenses = f
The equivalent focal length of a system of two lenses in contact is given as:
1 / f = 1 / f1 + 1 / f2
1 / f = 1 / 30 - 1 / 20 = 2 - 3 / 60 = - 1 / 60
So f = - 60 cm
Hence, the focal length of the combination of lenses is 60 cm. The negative sign indicates that the system of lenses acts as a diverging lens.
Q25 Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision? Ans: A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye- lens loses its ability of accommodation, the defect is called presbyopia.
Q26 A myopic person has been using spectacles of power −1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have happened. Ans: The power of the spectacles used by the myopic person, P = −1.0 D
Focal length of the spectacles, f = 1/P = 1/-1x10-2 = -100 cm
Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm. He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.
During old age, the person uses reading glasses of power, P' = +2 D
The ability of accommodation is lost in old age. This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.
Q27 A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected? Ans: In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is called astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.
Chapter 9 Ray Optics And Optical Instruments
You will get to learn about the optical instruments in use, spherical mirrors, and their activities in reflecting as well as refracting light, mirror formula, optical fibres, thin lens formula, etc in the ninth chapter of Class 12 Physics. This chapter also teaches you about microscopes and astronomical telescopes and their magnifying powers.
Download pdf of NCERT Solutions for Class Physics Chapter 9 Ray Optics And Optical Instruments
Exercise 1
Popular Questions of Class 12 Physics
- Q:-
What is the force between two small charged spheres having charges of 2 x 10-7 C and 3 x 10-7 C placed 30 cm apart in air?
- Q:-
An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.
- Q:-
A polythene piece rubbed with wool is found to have a negative charge of 3 × 10−7 C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
- Q:-
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
- Q:-
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10-12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
- Q:- A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
- Q:-
A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
- Q:-
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
- Q:-
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?
- Q:-
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Recently Viewed Questions of Class 12 Physics
- Q:-
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
- Q:-
Two charges -q and +q are located at points (0, 0, - a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
- Q:-
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
- Q:-
An electric dipole with dipole moment 4 × 10−9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104N C−1. Calculate the magnitude of the torque acting on the dipole.
- Q:-
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz - Q:-
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/103).
- Q:-
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
- Q:-
Determine the current in each branch of the network shown in figure
- Q:-
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
- Q:-
A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of west to 10°north of east. The magnetic meridian of the place happens to be 10º west of the geographic meridian. The earth's magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)
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