Chapter 9 Ray Optics and Optical Instruments Solutions
Question - 1 : - Asmall candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror ofradius of curvature 36 cm. At what distance from the mirror should a screen beplaced in order to obtain a sharp image? Describe the nature and size of theimage. If the candle is moved closer to the mirror, how would the screen haveto be moved?
Answer - 1 : -
Size of the candle, h= 2.5 cm
Image size = h’
Object distance, u= −27 cm
Radius of curvature of the concavemirror, R= −36 cm
Focallength of the concave mirror,
Image distance = v
The image distance can beobtained using the mirror formula:
Therefore, the screenshould be placed 54 cm away from the mirror to obtain a sharp image.
Themagnification of the image is given as:
The height of the candle’simage is 5 cm. The negative sign indicates that the image is inverted and real.
Ifthe candle is moved closer to the mirror, then the screen will have to be movedaway from the mirror in order to obtain the image.
A4.5 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 asthe needle is moved farther from the mirror.
Answer - 2 : -
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
Thevalue of v can be obtained using the mirror formula:Hence, the image of theneedle is 6.7 cm away from the mirror. Also, it is on the other side of themirror.
The image size is given bythe magnification formula:
Hence, magnification of the image, 
The height of the image is2.5 cm. The positive sign indicates that the image is erect, virtual, anddiminished.
Ifthe needle is moved farther from the mirror, the image will also move away fromthe mirror, and the size of the image will reduce gradually.
Atank is filled with water to a height of 12.5 cm. The apparent depth of aneedle lying at the bottom of the tank is measured by a microscope to be 9.4cm. What is the refractive index of water? If water is replaced by a liquid ofrefractive index 1.63 up to the same height, by what distance would themicroscope have to be moved to focus on the needle again?
Answer - 3 : -
Actual depth of the needle in water, h1 = 12.5 cm
Apparent depth of the needle in water, h2 = 9.4 cm
Refractive index of water = μ
The value of μcan be obtained asfollows:
Hence, the refractiveindex of water is about 1.33.
Wateris replaced by a liquid of refractive index, 
The actual depth of the needle remains thesame, but its apparent depth changes. Let y be the newapparent depth of the needle. Hence, we can write the relation:
Hence, the new apparent depth of the needleis 7.67 cm. It is less than h2. Therefore, to focus theneedle again, the microscope should be moved up.
∴Distanceby which the microscope should be moved up = 9.4 − 7.67
=1.73 cm
Question - 4 : - Figures 9.34(a) and (b)show refraction of a ray in air incident at 60° with the normal to a glass-airand water-air interface, respectively. Predict the angle of refraction in glasswhen the angle of incidence in water is 45º with the normal to a water-glassinterface [Fig. 9.34(c)].
Answer - 4 : -
As per the given figure,for the glass − air interface:
Angle of incidence, i = 60°
Angle of refraction, r =35°
The relative refractiveindex of glass with respect to air is given by Snell’s law as:
As per the given figure,for the air − water interface:
Angle of incidence, i = 60°
Angle of refraction, r =47°
Therelative refractive index of water with respect to air is given by Snell’s lawas:
Using (1) and (2), therelative refractive index of glass with respect to water can be obtained as:
The following figure showsthe situation involving the glass − water interface.
Angle of incidence, i = 45°
Angle of refraction = r
From Snell’s law, r can becalculated as:
Hence,the angle of refraction at the water − glass interface is 38.68°.
Question - 5 : - Asmall bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulbcan emerge out? Refractive index of water is 1.33. (Consider the bulb to be apoint source.)
Answer - 5 : -
Actual depth of the bulb in water, d1 = 80 cm = 0.8 m
Refractiveindex of water, The given situation isshown in the following figure:
Where,
i = Angle of incidence
r = Angle of refraction = 90°
Sincethe bulb is a point source, the emergent light can be considered as a circle ofradius,
Using Snell’ law, we canwrite the relation for the refractive index of water as:
Using the given figure, wehave the relation:
∴R = tan 48.75° × 0.8 = 0.91m
∴Area of the surface ofwater = πR2 = π (0.91)2 = 2.61 m2
Hence,the area of the surface of water through which the light from the bulb canemerge is approximately 2.61 m2.
A prismis made of glass of unknown refractive index. A parallel beam of light isincident on a face of the prism. The angle of minimum deviation is measured tobe 40°. What is the refractive index of the material of the prism? Therefracting angle of the prism is 60°. If the prism is placed in water(refractive index 1.33), predict the new angle of minimum deviation of aparallel beam of light.
Answer - 6 : - Angle of minimum deviation, 
= 40°
Angle of the prism, A = 60°
Refractive index of water, µ =1.33
Refractiveindex of the material of the prism = 
The angle of deviation is related to refractiveindex
as: Hence, the refractiveindex of the material of the prism is 1.532.
Sincethe prism is placed in water, let 
be the new angle ofminimum deviation for the same prism.The refractive index ofglass with respect to water is given by the relation:
Hence,the new minimum angle of deviation is 10.32°.
Question - 7 : - Double-convexlenses are to be manufactured from a glass of refractive index 1.55, with bothfaces of the same radius of curvature. What is the radius of curvature requiredif the focal length is to be 20 cm?
Answer - 7 : - Refractive index of glass,
Focal length of the double-convex lens, f =20 cm
Radius of curvature of one face of the lens= R1
Radius of curvature of the other face of thelens = R2
Radiusof curvature of the double-convex lens = RThe value of R can becalculated as:
Hence,the radius of curvature of the double-convex lens is 22 cm.
Abeam of light converges at a point P. Now a lens is placed in the path of theconvergent beam 12 cm from P. At what point does the beam converge if the lensis (a) a convex lens of focal length 20 cm, and (b) a concave lens of focallength 16 cm?
Answer - 8 : -
In the given situation,the object is virtual and the image formed is real.
Object distance, u = +12 cm
(a) Focal length of theconvex lens, f = 20 cm
Image distance = v
According to the lensformula, we have the relation:
Hence, the image is formed7.5 cm away from the lens, toward its right.
(b) Focal length of theconcave lens, f = −16 cm
Image distance = v
Accordingto the lens formula, we have the relation:
Hence,the image is formed 48 cm away from the lens, toward its right.
Question - 9 : - Anobject of size 3.0 cm is placed 14 cm in front of a concave lens of focallength 21 cm. Describe the image produced by the lens. What happens if theobject is moved further away from the lens?
Answer - 9 : -
Size of the object, h1 = 3 cm
Object distance, u = −14 cm
Focal length of the concave lens, f =−21 cm
Image distance = v
According to the lensformula, we have the relation:
Hence, the image is formedon the other side of the lens, 8.4 cm away from it. The negative sign showsthat the image is erect and virtual.
Themagnification of the image is given as:
Hence, the height of theimage is 1.8 cm.
Ifthe object is moved further away from the lens, then the virtual image willmove toward the focus of the lens, but not beyond it. The size of the imagewill decrease with the increase in the object distance.
Question - 10 : - Whatis the focal length of a convex lens of focal length 30 cm in contact with aconcave lens of focal length 20 cm? Is the system a converging or a diverginglens? Ignore thickness of the lenses.
Answer - 10 : -
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 focallength of a system of two lenses in contact is given as:
Hence,the focal length of the combination of lenses is 60 cm. The negative signindicates that the system of lenses acts as a diverging lens.