Conceptual Problems, Unit 6
Conceptual Problems, Unit 6
It includes the conceptual problems for ambitious, brilliant and curious students for competitions
Q.6.01.
The color of the surfaces of the bodies that can be impregnated with water becomes darker and richer after moistening. Explain the phenomenon.
Conceptual Problems, Unit 6
Solution:
Any surface’s color is determined with the help of the spectral makeup of the light beams it reflects.
A body’s surface has roughness and inconsistencies when it is dry. White light is randomly diffused when it is incident on the dry surface of the body. The white light that has been randomly diffused is superimposed with the light rays that correspond to the hue of the body. As a result, the body’s primary hue appears faded and less vibrant.
When a body’s surface is moistened with water, all of the surface imperfections are covered with water, preventing the white light incident on the moistened surface from being randomly diffused.
Conceptual Problems, Unit 6
Q.6.02.
Why is there summer in December and winter in June in the southern hemisphere?
Conceptual Problems (Unit 6)
Solution:
The angle of incidence of sun rays is minimum in December and maximum in June in the southern hemisphere of the earth. Therefore, it follows that the intensity of heat radiation on the surface of the earth will reach its maximum in December and its minimum in June. Hence, it is summer in December and winter in June in the southern hemisphere.
Q.6.03.
A convex and concave mirror, each of focal length f, is placed at a distance of 4 f apart, as shown in Fig. 6.01. At what point on the principal axis of the two mirrors should a point source of light be placed for the rays to converge at the same point after being reflected first from the convex mirror and then from the concave mirror? Where will the rays meet if they are first reflected from the concave mirror?
Conceptual Problems, Unit 6
Solution:
The convex and concave mirrors, each having a focal length f, are placed 4 f apart, as shown in the figure.
Suppose that if the source of light is placed at point O, the rays of light coverage will be at the same point after being reflected first from the convex mirror and then from the concave mirror. Firstly, the convex mirror will be from the virtual image I1. Image I1, then, acts as the real object for the concave mirror, and its final image is produced at point O, as shown in Figure 6.01
For reflection from the convex mirror:
If u1, v1, and f1 denote the object distance, image distance, and focal length, respectively, then (1/u1) + (1/v1) = 1/f1 Here, f1 = + f (convex mirror )
Suppose that the object is placed at a distance x from the pole of the convex mirror. Then, u1 = -x. Therefore, the mirror formula becomes
1/(-x) + (1/v1) = 1/f or (1/v1) = (1/x) + (1/f) = (f + x) /f x
or v1 = f x / (f + x)
For reflection from the concave mirror:
If u2, v2, and f2 denote the object distance ( I1 acts as the real object ) , image distance, and focal length, respectively, then
1/u2 + 1/v2 = 1/ f2 In this case, f2 = – f (concave mirror), and u2 = -(v1 + 4 f) = – [f x / (f + x) + 4 f]
= – {f x + 4 f (f + x) } / (f + x) = – [(5 x + 4 f ) f ] / (f + x)
Therefore, the mirror formula becomes
(f + x) / [(5 x + 4 f ) f ] + 1 /v2 = 1/ (-f)
Or 1/v2 = (f + x) / [(5 x + 4 f ) f ] – 1/f = – [(f + x) – [(5 x + 4 f)]] / [(5 x + 4 f ) f ]
= – (3 f + 4x) / [(5 x + 4 f ) f ]
Or v2 = -[(5 x + 4 f ) f] / (3 f + 4 x)
Now, P1 P2 = OP1 + OP2 = u1 + v2 = 4 f
Therefore, x+{(5 x + 4 f ) f/ (3 f + 4 x)} = 4 f
Or {x(3 f + 4 x ) +(5 x + 4 f ) f} / (3 f + 4 x) = 4 f
Or x(3 f + 4 x ) +(5 x + 4 f ) f = 4 f (3 f + 4 x)
Or 4 x2 – 8 f x – 8 f2 = 0
Or x2 – 2 f x – 2 f2 = 0
Therefore, x = 2f ±{ (- 2 f)2 – 4⨯1⨯(- 2 f2)}1/2 / 2
= {2f ± (4f2+ 8 f2) 1/2}/2 = f ± = (1± √3) f
Since x cannot be negative,
x = (1 +√3)f (from convex mirror )
If the initial path of light rays is reversed, the position of the point where the ray finally meets will remain the same. It is because of the principle of the reversibility of light.
Conceptual Problems (Unit 6)
Q.6.04.
A concave mirror of focal length 25 cm forms the real image of a point object at a distance O lying on its principal axis at a distance of 50 cm from the mirror. The mirror is cut into two halves and drawn at a distance of 1 cm apart in the direction perpendicular to the optical axis, as shown in Fig. 6.02. How will the two halves of the concave mirror produce the image of the point object O?
Conceptual Problems (Unit 6)
Solution:
Fig. 6.02 shows a concave mirror whose pole is P and focus is F (PF = 25 cm). A point object O is placed on the principal axis PX at a distance OP = 50 cm.
When the mirror is cut into two halves and the two halves are drawn 1 cm apart, object O lies at a distance OM = ON = 0.5 cm from the principal axes P1 X1 and P2 X2 of the two halves of the mirror [Fig. 6.03]
For the upper half of the concave mirror, object O lies at a distance of 2 f (= 50 cm) from it but is displaced through a distance of 0.5 cm below its principal axis P1 X1. Therefore, the image O1 of the point object O will be formed at a distance of 2 f but displaced through a distance of 0.5 cm above its principal axis.
Similarly, the lower half of the concave mirror will produce the image O2 of the point object O at a distance of 0.5 cm below its principal axis P2 X2. The distance between the images O1 and O2 will be 2 cm. Hence, when the concave mirror is cut into two halves and the two halves are drawn 1 cm apart, the two halves will produce two-point images at a distance of 2 cm from each other and at a distance of 50 cm from the concave mirror.
Conceptual Problems (Unit 6)
Q.6.05
A fish swims in a glass tank. A person whose eye is above the level of water sees two fish. Draw a ray diagram to illustrate this.
Conceptual Problems (Unit 6)
Solution:
Fig. 6.04 shows a fish O inside the water. A person whose eye is above the seawater sees two fish, one due to refraction of the light taking place from the surface of the water and the other due to refraction from the vertical wall of the glass tank.
Conceptual Problems (Unit 6)
Thus, image I1 of the fish is seen due to the refraction of the light at point A on the surface of the water, and image I2 is due to refraction at point B on the vertical wall of the glass tank. This is how a person whose eye is above the level of water sees two fish.
Conceptual Problems (Unit 6)
Q.6.06
The moon moves across the sky at approximately 0.5 miles per hour.
(a) When it passes in front of a star, what do you see? Discuss
(b) What would you see if the moon had an atmosphere like the Earth?
Conceptual Problems (Unit 6)
Solution:
(a) Light travels along a straight path from the star to the earth through a vacuum. Therefore, until the solid edge of the moon cuts off the light, the star will not be seen. It will remain fixed compared to other stars, and its brightness will also remain constant.
Conceptual Problems (Unit 6)
(b) If the moon had an atmosphere (shown by the shaded circle around the moon) and the light from the star S traversed its path through it, then light from the star would be refracted, as shown in Fig. 6.05, and the star would seem to shift from S to S’ Thus, as the moon moves towards the line of sight of the star, the star seems to move ahead of the moon. Due to the scattering of the blue part of the visible spectrum from the moon’s atmosphere, it will appear red and dim before the solid edge of the moon cuts off its light.
Conceptual Problems (Unit 6)
Q.6.07.
The greatest thickness of a plano-convex lens, when viewed normally through its plane surface, appears to be 8/300 m and when viewed normally through its curved surface, appears to be 16/500 m. If the actual thickness is 0.04 m, find its (1) refractive index, (2) radius of curvature, and (3) focal length.
Conceptual Problems (Unit 6)
Solution:
When the plano-convex lens is viewed through its plane surface, the ray of light from point P on the convex surface reaches the eye after suffering refraction at the plane surface of the plano-convex lens, as shown in Fig. 6.
Due to the normal shift, the point P appears to be at I. Thus, the actual thickness of the lens, X = PQ = 0.04 m and the apparent thickness of the lens,
Y = PI = 8/ 300 m
If µ is the refractive index of the material of the lens, then
When the plano-convex lens is viewed through its curved surface, In this case, the ray of light from point Q on the plane surface reaches the eye after suffering refraction at the convex spherical surface, whose center of curvature is C. The convex spherical surface forms the virtual image I of the object Q, as shown in Fig. 6.07.
If u(= P Q) and v (= P I) are the distance of the object and image from the pole P of the spherical surface, then according to the new cartesian sign conventions,
u = – 0.04 m and v = – 16 / 500 m
Let R (= P C) be the radius of curvature of the convex spherical surface of the lens. As the incident ray travels in the denser medium, we have
According to the lens maker’s formula, we have
1/ f = (µ – 1) {1/ R1 – 1/R2}
For the given plano-convex lens,
R1 = ∞ and R2 = – 0.08 m = R
Therefore, 1/f = (1.5 -1) {1/∞ – 1/ -0.08}
= 0.5 ⨯{0+ 1/ 0.08 } = 1/0.16
Or f = 0.16 m
Q.6.08.
What is the minimum distance between an object and the real image formed by a convex lens?
Conceptual Problems (Unit 6)
Solution:
For a convex lens, we have
-1/ u + 1/ v = 1/f ……(1)
Suppose that an object is placed at a distance x from the convex lens, and its real image is formed on the other side of the lens at a distance v from it [Fig. 6.08].
Now, u = – x ; v = + v ; f = + f
Therefore, equation (1) becomes
-1/ -x + 1/ +v = 1/ +f
Or 1/ v = 1/f – 1/x = (x – f) / f x
Or v = f x / x – f …….(2)
If D is the distance between the object and the image, then
v = D – x (3)
From equations (2) and (3), we have
D – x = f x / (x – f) ……..(4)
Differentiating both sides of equation (4) w.r.t. x, we have
dD/dx = 1 +f/(x – f) + f x (-1) (x -f)-2
= 1 + f/ (x – f) – f x/(x – f)2
For D to be minimum,
dD/dx = 0
That is 1 + f/ (x – f) – f x/(x – f)2 = 0
Or x /x – f = f x / (x -f)2 = f / x – f = 1
Or x = 2f
In equation 4, substituting x = 2 f, we have
D = 2 f +( f ⨯2 f)/ 2 f – f = 2 f + 2 f = 4 f
Q.6.09.
A convex and concave lens having focal lengths of 30 cm and 10 cm, respectively, are placed at a distance of 20 cm from each other. Where should a source of light be placed so that the system of the two lenses will produce a beam of parallel rays?
Conceptual Problems (Unit 6)
Solution:
Suppose that convex lens L1 (f1 = 30 cm) and concave lens L2 (f2 = 10 cm) are placed at a distance of 20 cm apart, as shown in Fig. 6.09.
Suppose that the beam of light, after falling on the convex lens, converges at point P. As the beam of light is rendered parallel due to the concave lens, the point p is the focus of the concave lens. Therefore,
C1P = C1 C2 + C2P = 20 + 10 = 30 cm
Since the focal length of the convex lens is also 30 cm, point P is the focus of the convex lens. Thus, the beam of light, after falling on the convex lens, converges at its focus. Hence, a parallel beam of light is falling on the convex lens, which requires that the source of light be situated at infinity.
Q.6.10
An ant is approaching a convex lens with a uniform speed up to the first focus. How would the speed of the image of the ant change due to the lens change?
Conceptual Problems (Unit 6)
Solution:
From the lens equation, we have
-1/u + 1/v = 1/f
Or v = u f /u + f
Differentiating both sides, we have
dv/ dt = u +f ⨯ f du/ dt – u f ⨯[du/dt + 0
Or dv/dt = f2 / (u + f)2 ……….(1)
If u and v are the distance of the ant and its image from the optical center of the lens, du/dt and dv/dt represent the speed of the ant and its image, respectively.
Now, for a convex lens, f is positive and u is negative. Therefore, the factor
f2 /(u + f)2 > 1
Then, from equation (1), it follows that DV/dt > du/dt, i.e., the image of the ant moves with a speed greater than that of the ant.
Q.6.11
A beam of white light passing through a hollow prism has no spectrum. Explain.
Conceptual Problems (Unit 6)
Solution:
Consider a hollow prism, as shown in Fig. 6.10. Here, ABC and A’ B ‘C represent the outer and inner surfaces of the prism. The three faces of the prism simply act as three glass slabs with parallel faces.
When a ray of light is incident on the face AB, it will pass through a glass slab formed between the surfaces AB and A*B*. A ray of light will not disperse, but it will get laterally displaced. A similar effect will take place when the ray falls on the inner surfaces of A* and C*. Thus, the ray of light will come out of the prism without being dispersed but getting laterally displaced.
Q.6.12
Artificial satellites can often be seen as bright objects high in the sky long after sunset. What must be the minimum altitude of a satellite moving above the earth’s equator to be still visible directly overhead two hours after sunset?
Conceptual Problems (Unit 6)
Solution:
At sunset, the sunlight will be tangent to the earth’s surface. Further, in two hours, the Earth will rotate by 30°. Therefore, the satellite will come in the path of the sunlight if its orbit is at a distance BS = h (say) above the earth’s equator [Fig. 6.11]
If OA = OB = R is the radius of the earth, then from the right-angled triangle OAB, we have
Cos 30̯ = OA/OS = OA/OB + BS = R/R + H
or h = R/cos 30o – R = R/0.866 – R = 0.155 R
Q.6.13
Occasionally, one sees an aircraft in the sky shortly after sunset or shortly before sunrise and is surprised to see that it appears to be very bright, more like a star or planet than like an aircraft as seen in daylight Explain, why?
Conceptual Problems (Unit 6)
Solution:
Shortly after sunset or before sunrise, an observer on the surface of the earth does not receive sunlight directly from the sun. The observer receives the light reflected from the aircraft, as shown in Fig. 6.12.
In the reflected sunlight, the spacecraft appears bright against the darkening background of the early morning or evening sky.
Q.6.14
Is the human eye sensitive to all the wavelengths in the visible region? How is the sensitivity of the human eye measured at a particular wavelength?
Conceptual Problems (Unit 6)
Solution:
The sensitivity of the human eye is not the same for all the wavelengths (colors) of visible light. It is found to be the most sensitive, corresponding to a wavelength of 5,500 Å in the yellow region. The sensitivity of the human eye to a particular wavelength of light is measured as the sensation of sight produced for a given power per unit area of the light of that wavelength incident on the eye.
Q.6.15
When observed from the earth, the angular diameter of the solar disc is 32′. Determine the diameter of the image of the sun formed due to a convex lens of focal length 25 cm.
Conceptual Problems (Unit 6)
Solution:
Here, f = 25 cm
And angular diameter of the sun,
Ѳ = 32′ = ꙥ/180 ⨯32/60 rad
The image of the solar disc will be formed in the focal plane of the lens [Fig. 6.13].
At the optical center of the lens, the diameter (D) of the image of the solar disc will subtend an angle equal to the angular diameter of the sun. therefore,
D = f Ѳ
= 25⨯ ꙥ/ 180 ⨯ 32/60 = 0.233 cm
Q.6.16
A refracting telescope has an objective of a focal length of 1 m and an eye-piece of a focal length of 20 cm. A real image of the sun, 10 cm in diameter, is formed on a screen 24 cm from the eyepiece. What angle does the sun subtend at the objective?
Conceptual Problems (Unit 6)
Solution:
Here, f0 = 1 m = 100 cm; fe = 20 cm; D = 24 cm;
A”B” = 10 cm
If A”B” is the size of the image formed by the object lens, then the angle formed by the sun at the objective will be tan α = A’B’/f0
To find A’B’: As shown in Fig. 4.10 (chapter), A’B’ acts as an object for the eyepiece, and its virtual image A’B’ is formed at a distance of 24 cm. Therefore, A”B”/A’B’ = 1+ D/fo Or 10/A’B’ = 1+ 24/20 = 44/20 Or A’B’ = 50/11 cm. Therefore,
tanα = 50/11 x 100 = 0.0455. Since the angle α is small, therefore, α = tanα = 0.0455 radian
Q.6.17
Can you suggest reasons why no interference is seen when light reflects from the two surfaces of a window plane?
Conceptual Problems (Unit 6)
Solution:
Suppose that the glass window pane is viewed in white light, and the glass pane is say 0.2 cm thick. For light of wavelength 6,000 Å, the path difference between the light reflected from the two surfaces will be
2 µ t = 2⨯1.5 ⨯0.2 = 0.6 cm.
This path difference of 0.6 cm is equivalent to 10,000 wavelengths. In a reflected system, the light will interfere destructively and will appear dark. It can be calculated that, corresponding to light of wavelength 5,999.7 Å, the path difference will be 10,000.5 wavelengths. Lights of this wavelength will interfere constructively in the reflected system. Thus, many wavelengths evenly distributed over the entire spectrum would be reflected, and the net result would be not colored interference fringes but white reflected light.
If monochromatic light were used, we might obtain interference fringes. However, the surface of ordinary window glass is too far from being optically plane to produce interference fringes.
Q.6.18
One method of producing two in-phase point sources of light for interference is to form two images of a point source using the two halves of a lens, which have been split along a diameter [Fig. 6.14].
(a) How must the two halves of a 5 cm focal length lens be placed to form two real image sources 0.24 cm apart and 20 cm from the source?
(b) A screen is placed perpendicular to the principal axis and 80 cm from the image sources. What is the width of the central maxima formed on the screen if the wavelength of the light used is 5.4 ⨯ 10-5 cm?
Solution:
(a) As one lens, the two halves will form a single point on the principal axis of a point source located on the principal axis. As the two halves of the lens are separated, each half will produce an image with its own principal axis. If the two halves of the lens are separated through a distance of x, the principal axis of each half is now x/2 from the line joining the source S to the point O on the screen. The source will then be at a distance of x/2 off the principal axis of each half of the lens.
Since the two images, S1 and S2, are produced at a distance of 20 cm from the source S, u = v =20 cm. As the focal length of the lens is 5 cm, the images of S1 and S2 are to be produced at a distance of 20 cm from the source, if u = v = 20/2 = 10 cm, i.e., the two halves should be placed at a distance of 10 cm from the source. As such, the image produced due to each half of the lens will be x/2 on the other side of its principal axis. If d is the distance between the two imagesS1 and S2, then
d = 4⨯(x/2) =2 x
Here, d = 0.24 cm
Therefore, 2 x = 0.24 cm
or x = 0.12 cm
(b ) Now, the width of the central maxima,
β0 = D λ/ d
Here, D = 80 cm, d = 0.24 cm and λ = 5.4 ⨯ 10-5 cm
Therefore, β0 = (80⨯ 5.4 ⨯ 10-5 ) 0.24 = 1.8 ⨯10-2 cm
Q.6.19
What will happen to the interference pattern in Young’s experiment if the source is not exactly on the center line between the slits?
Solution:
In the normal position of the source w.r.t. the double slits [Fig. 6.15], the central bright fringe is formed at the point O.
It is because the paths traversed by the light along SAO and SBO are equal. When the source is moved upwards, the central fringe will occur at the point O* on the screen [Fig. 6.16], where the paths of light S*AO* and S*BO* are equal.
Therefore, when the source of light is not exactly on the center line and is moved upwards, the central bright fringe will move downwards.
Q.6.20
If a thin soap film is arranged vertically, the colored horizontal interference bands move downward and, at the same time, change their width. After some time, a rapidly growing dark spot appears at the top of the film, which bursts shortly afterward. Explain, why?
Solution:
In a thin soap film, the colored bands are produced due to the interference of light waves reflected from the upper and lower surfaces of the soap film. The path difference between the light waves reflected from the two surfaces depends on the thickness of the film at the point of reflection and the wavelength of the light waves. Therefore, each of the horizonal interference bands corresponds to some definite thickness of the film.
In a soap film arranged vertically, water flows gradually from the upper part to the lower part of the film. Due to this, the lower part gradually grows thicker while the upper part becomes thinner. The points corresponding to some definite thickness are gradually displaced, and the respective interference bands move downward with them. After some time, the thickness of the film in the upper part becomes equal to one-quarter of the shortest wavelength of the incident light. At this point of the film, the light waves of this wavelength reflected from the two surfaces of the film will interfere destructively, and the film will appear dark at that point in the reflected light.
Q.6.21
Lenses are often coated with a thin film to reduce the intensity of reflected light.
(a) If the index of refraction of the coating is 1.3, what is the smallest thickness that will give a minimum reflection of yellow light (λ = 5,800 Å)? (b) Such lenses often show a faint purple color from reflected light. Why?
Solution:
When light falls on a lens (µ = 1.5) coated with a thin film (µ = 1.3), a part of the light gets partially reflected from the film and partially from the lens. The path difference between the light rays reflected from the film and the glass depends on the thickness of the film. Since in both cases, reflection takes place in a denser medium, no extra path difference (or phase difference) is created. A film of thickness λ/4 will give a total path difference of λ/2, and the two reflected light waves will interfere destructively. In this manner, a thin film coated on the lens helps to reduce the intensity of the reflected light. The wavelength of the yellow light inside the film,
λ = λair/ λfilm = 5,800/ 1.3 = 4,462 Å
The minimum reflection of the yellow light thickness of the film is given by
t = λ / 4 = 4,462/4 = 1,115.5 Å
(b) If the film is of the proper thickness to produce destructive interference yellow light, red light, and violet light will be somewhat reflected. The reflected light will therefore have a reddish purple tinge.
