Focal length of the convex lens, f1 = 30 cm Focal length of the concave lens, f2 = -20 cm Distance between the two lenses, d = 8.0 cm Ans (a). When the parallel beam of light is incident on the convex lens first: According to the lens formula, we have: 1/v₁ -1/u₁= 1/f1 Where, u1 = Object distance =Read more
Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = -20 cm
Distance between the two lenses, d = 8.0 cm
Ans (a).
When the parallel beam of light is incident on the convex lens first:
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
1/v2 -1/u2 = 1/f2
Where, u2 = Object distance = (30 – d) = 30 – 8 = 22 cm
v2 = Image distance
1/v2 = 1/22-1/20 = (10-11)/220 = -1/220
Therefore, v2 = -220 cm
The parallel incident beam appears to diverge from a point that is (220-d/2 = 220 — 4) i.e. 216 cm from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula,
We have:
1/v2 – 1/u2 = 1/f2
=> 1/v2 = 1/f2 + 1/u2
Where, u2 = Object distance = – ∝
v2 = Image distance
1/v2 = 1/(-20 ) + 1/(- ∝) = -1/20
v2 = -20cm
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
Hence, the parallel incident beam appear to diverge from a point that is (420 – 4) 416 cm from the left of the centre of the combination of the two lenses.
The answer does depend on the side of the combination at which the parallel beam of light is incident The notion of effective focal length does not seem to be useful for this combination.
Ans (b).
Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens, u1 = -40 cm
|u1| = 40 cm
According to the lens formula:
1/v1-1/u1=1/f1
Where,
v1= Image distance
1/v1 = 1/30 + 1/(-40) =( 4-3)/120 =1/120
Therefore ,v1 = 120 cm
Magnification, m = v1/|u1| = 120/40 = 3
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens. According to the lens formula:
1/v2 – 1/u2 = 1/f2
Where, u2 = Object distance = + (120 – 8) = 112 cm
Distance between the image (screen) and the object, D = 90 cm Distance between two locations of the convex lens, d = 20 cm Focal length of the lens = f Focal length is related to d and D as: f2 = (D2-d2)/4D = [(90)2 - (20)2]/ (4 x 90) = 770 /36 = 21.39 cm Therefore, the focal length of the convexRead more
Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
f2 = (D2-d2)/4D
= [(90)2 – (20)2]/ (4 x 90) = 770 /36 = 21.39 cm
Therefore, the focal length of the convex lens is 21.39 cm
Distance between the object and the image, d = 3 m Maximum focal length of the convex lens = fmax For real images, the maximum focal length is given as: fmax = d/4 =3/4 =0.75 m Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
Distance between the object and the image, d = 3 m
Maximum focal length of the convex lens = fmax
For real images, the maximum focal length is given as:
fmax = d/4 =3/4 =0.75 m
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
Ans (a). Yes Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed. Ans (b). NoRead more
Ans (a).
Yes
Plane and convex mirrors can produce real images as well. If the object is virtual,
i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.
Ans (b).
No
A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.
Ans (c).
The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.
Ans (d).
Yes; Decrease
The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.
Ans (e).
Yes
The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.
(a) Refractive index of the glass fibre, μ₁ = 1.68 Refractive index of the outer covering of the pipe, μ2 = 1.44 Angle of incidence = i Angle of refraction = r Angle of incidence at the interface = i’ The refractive index (μ) of the inner core - outer core interface is given as: μ =μ₂/μ₁= 1/ sini' sRead more
(a) Refractive index of the glass fibre, μ₁ = 1.68
Refractive index of the outer covering of the pipe, μ2 = 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (μ) of the inner core – outer core interface is given as:
μ =μ₂/μ₁= 1/ sini’
sin i’ = μ₁/μ₂ = 1.44/1.68 = 0.8571
Therefore, i’= 59°
For the critical angle, total internal reflection (TIR) takes place only when i > i’ ,
i.e., i > 59º
Maximum angle of reflection,
rmax = 90º – i’ = 90º – 59º = 31º
Let , imax be the maximum angle of incidence.
The refractive index at the air – glass interface, μ₁= 1.68
We have the relation for the maximum angles of incidence and reflection as:
μ₁ = sin imax/ sin rmax
sin imax = μ₁ sin rmax
= I.68 sin31°
= 1.68×0.5150 = 0.8652
Therefore, imax = sin⁻¹ 0.8652 ≈ 60°
Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.
Ans (b).
If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ₁ = Refractive index of air = 1
For the angle of incidence i = 90°,
We can write Snell’s law at the air – pipe interface as:
sini/sin r = μ2 = 1.68
sin r = sin 90° /1.68 = 1/1.68
r = sin⁻¹ (0.5952)
= 36.5°
Therefore, i’ = 90° – 36.5° = 53.5°
Since i’ > r, all incident rays will suffer total internal reflection.
(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Focal length of the convex lens, f1 = 30 cm Focal length of the concave lens, f2 = -20 cm Distance between the two lenses, d = 8.0 cm Ans (a). When the parallel beam of light is incident on the convex lens first: According to the lens formula, we have: 1/v₁ -1/u₁= 1/f1 Where, u1 = Object distance =Read more
Focal length of the convex lens, f1 = 30 cm
Focal length of the concave lens, f2 = -20 cm
Distance between the two lenses, d = 8.0 cm
Ans (a).
When the parallel beam of light is incident on the convex lens first:
According to the lens formula, we have:
1/v₁ -1/u₁= 1/f1
Where, u1 = Object distance = ∝ and v1 = Image distance
1/v₁ -1/∝= 1/30
=> v1 = 30 cm
The image will act as a virtual object for the concave lens.
Applying lens formula to the concave lens, we have:
1/v2 -1/u2 = 1/f2
Where, u2 = Object distance = (30 – d) = 30 – 8 = 22 cm
v2 = Image distance
1/v2 = 1/22-1/20 = (10-11)/220 = -1/220
Therefore, v2 = -220 cm
The parallel incident beam appears to diverge from a point that is (220-d/2 = 220 — 4) i.e. 216 cm from the centre of the combination of the two lenses.
(ii) When the parallel beam of light is incident, from the left, on the concave lens first:
According to the lens formula,
We have:
1/v2 – 1/u2 = 1/f2
=> 1/v2 = 1/f2 + 1/u2
Where, u2 = Object distance = – ∝
v2 = Image distance
1/v2 = 1/(-20 ) + 1/(- ∝) = -1/20
v2 = -20cm
The image will act as a real object for the convex lens.
Applying lens formula to the convex lens, we have:
1/v1 -1/u1 = 1/f1
Where, u1 = Object distance = – (20 + d) = – (20 + 8) = -28 cm
v1 = Image distance
1/v1 = 1/30 + 1/(-28) = (14-15 )/420 = -1/420
v2 = -420 cm
Hence, the parallel incident beam appear to diverge from a point that is (420 – 4) 416 cm from the left of the centre of the combination of the two lenses.
The answer does depend on the side of the combination at which the parallel beam of light is incident The notion of effective focal length does not seem to be useful for this combination.
Ans (b).
Height of the image, h1 = 1.5 cm
Object distance from the side of the convex lens, u1 = -40 cm
|u1| = 40 cm
According to the lens formula:
1/v1-1/u1=1/f1
Where,
v1= Image distance
1/v1 = 1/30 + 1/(-40) =( 4-3)/120 =1/120
Therefore ,v1 = 120 cm
Magnification, m = v1/|u1| = 120/40 = 3
Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens. According to the lens formula:
1/v2 – 1/u2 = 1/f2
Where, u2 = Object distance = + (120 – 8) = 112 cm
v2 = image distance
1/v2 = 1/(-20) + 1/112 = ( -112 + 20 )/2240 = -92/2240
Therefore , v2 = -2240/92 cm
Magnification m’ = |v2/u2| = (2240/92) x (1/112 ) = 20/92
Hence ,the magnitude due to the concave lens is 20/92
The magnification produced by the combination of the two lenses is calculated as:
m x m’ = 3 x 20 /92 = 60/92 = 0.652
The magnification of the combination is given as :
h2/h1 = 0.652
=> h2 = 0.652 x h1
Where,h1 = Object size = 1.5cm
h2 = size of the image
Therefore, h2 = 0.652 x 1.5 = 0.98 cm
Hence, the height of the image is 0.98 cm
See lessA screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Distance between the image (screen) and the object, D = 90 cm Distance between two locations of the convex lens, d = 20 cm Focal length of the lens = f Focal length is related to d and D as: f2 = (D2-d2)/4D = [(90)2 - (20)2]/ (4 x 90) = 770 /36 = 21.39 cm Therefore, the focal length of the convexRead more
Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
f2 = (D2-d2)/4D
= [(90)2 – (20)2]/ (4 x 90) = 770 /36 = 21.39 cm
Therefore, the focal length of the convex lens is 21.39 cm
See lessThe image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
Distance between the object and the image, d = 3 m Maximum focal length of the convex lens = fmax For real images, the maximum focal length is given as: fmax = d/4 =3/4 =0.75 m Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
Distance between the object and the image, d = 3 m
Maximum focal length of the convex lens = fmax
For real images, the maximum focal length is given as:
fmax = d/4 =3/4 =0.75 m
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.
See lessAnswer the following questions: (a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain. (b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction? (c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is? (d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease? (e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
Ans (a). Yes Plane and convex mirrors can produce real images as well. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed. Ans (b). NoRead more
Ans (a).
Yes
Plane and convex mirrors can produce real images as well. If the object is virtual,
i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.
Ans (b).
No
A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.
Ans (c).
The diver is in the water and the fisherman is on land (i.e., in air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are travelling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.
Ans (d).
Yes; Decrease
The apparent depth of a tank of water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near-normal viewing.
Ans (e).
Yes
The refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.
See less(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b) What is the answer if there is no outer covering of the pipe?
(a) Refractive index of the glass fibre, μ₁ = 1.68 Refractive index of the outer covering of the pipe, μ2 = 1.44 Angle of incidence = i Angle of refraction = r Angle of incidence at the interface = i’ The refractive index (μ) of the inner core - outer core interface is given as: μ =μ₂/μ₁= 1/ sini' sRead more
(a) Refractive index of the glass fibre, μ₁ = 1.68
Refractive index of the outer covering of the pipe, μ2 = 1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’
The refractive index (μ) of the inner core – outer core interface is given as:
μ =μ₂/μ₁= 1/ sini’
sin i’ = μ₁/μ₂ = 1.44/1.68 = 0.8571
Therefore, i’= 59°
For the critical angle, total internal reflection (TIR) takes place only when i > i’ ,
i.e., i > 59º
Maximum angle of reflection,
rmax = 90º – i’ = 90º – 59º = 31º
Let , imax be the maximum angle of incidence.
The refractive index at the air – glass interface, μ₁= 1.68
We have the relation for the maximum angles of incidence and reflection as:
μ₁ = sin imax/ sin rmax
sin imax = μ₁ sin rmax
= I.68 sin31°
= 1.68×0.5150 = 0.8652
Therefore, imax = sin⁻¹ 0.8652 ≈ 60°
Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.
Ans (b).
If the outer covering of the pipe is not present, then:
Refractive index of the outer pipe, μ₁ = Refractive index of air = 1
For the angle of incidence i = 90°,
We can write Snell’s law at the air – pipe interface as:
sini/sin r = μ2 = 1.68
sin r = sin 90° /1.68 = 1/1.68
r = sin⁻¹ (0.5952)
= 36.5°
Therefore, i’ = 90° – 36.5° = 53.5°
Since i’ > r, all incident rays will suffer total internal reflection.
See less