Actual depth of the pin, d = 15 cm Apparent depth of the pin = d' Refractive index of glass,μ = 1.5 Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e. μ = d/d' Therefore ,d' = d/μ = 15/1.5 = 10 cm The distance at which the pin appears to be raised = d' - d =Read more
Actual depth of the pin, d = 15 cm
Apparent depth of the pin = d’
Refractive index of glass,μ = 1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
μ = d/d’
Therefore ,d’ = d/μ
= 15/1.5 = 10 cm
The distance at which the pin appears to be raised = d’ – d = 15 – 10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.
(a) For a concave mirror, the focal length (f) is negative. Therefore ,f < 0 When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore u < 0 For image distance v, we can write the lens formula as: 1/v - 1/u = 1/f 1/v = 1/f - 1/u------------------Read more
(a) For a concave mirror, the focal length (f) is negative. Therefore ,f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore u < 0 For image distance v, we can write the lens formula as:
1/v – 1/u = 1/f
1/v = 1/f – 1/u——————- Eq-1
The object lies between f and 2f.
However , 2f <u <f (because u and f are negative)
1/2f > 1/u > 1/f
-1/2f < -1/u <-1/f
1/f -1/2f <1/f -1/u<0 —————-Eq-2
Using equation (1), we get:
1/2f < 1/v < 0
Therefore, 1/v is negative, i.e., v is negative.
1/2f < 1/v => 2F > v
=> -v > -2f ,therefore, the image lies beyond 2f.
Ans (b).
For a convex mirror, the focal length (f) is positive. Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore, u < 0 For image distance v, we have the mirror formula:
1/v +1/u = 1/f
1/v = 1/f -1/u
Using equation (2), we can conclude that:
1/v<o
0r v>0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
Ans (c).
For a convex mirror, the focal length (f) is positive. f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore, u < 0 For image distance v, we have the mirror formula:
1/v + 1/u = 1/f
=> 1/v = 1/f -1/u
But we have u < 0
Therefore , 1/v > 1/f
or v <f
Hence, the image formed is diminished and is located between the focus (f) and the pole.
Ans (d).
For a concave mirror, the focal length (f) is negative. Therefore, f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore, u < 0 It is placed between the focus (f) and the pole.
Therefore ,f > u> 0
1/f < 1/u < 0
1/f -1/u < 0
For image distance v, we have the mirror formula:
1/v + 1/u = 1/f
1/v = 1/f -1/u
Therefore ,1/v < 0
v >0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write: 1/u > 1/v => v > u
Magnification, m = v/u > 1 Hence, the formed image is enlarged.
Focal length of the objective lens, fo = 15 m = 15 x 10² cm and focal length of the eyepiece, fe = 1.0 cm Ans (a). The angular magnification of a telescope is given as: α = fo/fe = 15 x 10²/1.0 = 1500 Hence, the angular magnification of the given refracting telescope is 1500. Ans (b). Diameter of tRead more
Focal length of the objective lens, fo = 15 m = 15 x 10² cm and focal length of the eyepiece, fe = 1.0 cm
Ans (a).
The angular magnification of a telescope is given as:
α = fo/fe = 15 x 10²/1.0 = 1500
Hence, the angular magnification of the given refracting telescope is 1500.
Ans (b).
Diameter of the moon, d = 3.48 x 10⁶ m and radius of the lunar orbit, ro = 3.8 x 10⁸ m Let d’ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
d/r0 = d’/f0
(3.48 x 10⁶)/(3.8 x 10⁸) = d’/15
d’= (3.48/3.8) x 10⁻² x 15
= 13.74 x 10⁻² m = 13.74 cm
Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm
Focal length of the objective lens, f0 = 144 cm Focal length of the eyepiece, fe = 6.0 cm The magnifying power of the telescope is given as: m = f0 /fe = 144/6 = 24 The separation between the objective lens and the eyepiece is calculated as: f0 + fe = 144 + 6 = 150 cm Hence, the magnifying poweRead more
Focal length of the objective lens, f0 = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as:
m = f0 /fe = 144/6 = 24
The separation between the objective lens and the eyepiece is calculated as:
f0 + fe = 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
Focal length of the objective lens, f0 = 8 mm = 0.8 cm Focal length of the eyepiece, fe = 2.5 cm Object distance for the objective lens, u0 = -9.0 mm = -0.9 cm Least distance of distant vision, d = 25 cm Image distance for the eyepiece, ve = -d = -25 cm Object distance for the eyepiece = ue Using thRead more
Focal length of the objective lens, f0 = 8 mm = 0.8 cm
Focal length of the eyepiece, fe = 2.5 cm
Object distance for the objective lens, u0 = -9.0 mm = -0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, ve = -d = -25 cm
Object distance for the eyepiece = ue
Using the lens formula, we can obtain the value of ue as:
1/ve – 1/ue = 1/fe
1/ue = 1/fe –1/ve
= 1/-25 – 1/2.5 = (-1-10)/25 = -11/25
Therefor, ue = – 25/11 = – 2.27
We can also obtain the value of the image distance for the objective lens ( v0) using the lens formula.
1/v0 -1/u0 = 1/f0
1/v0 = 1/f0+ 1/u0
= 1/0.8 -1/0.9 = (0.9 -0.8)/ 0.72 = 0.1/0.72
Therefore, v0 = 7.2 cm
The distance between the objective lens and the eyepiece
= |ue| + v0 = 2.27 + 7.2 = 9.47 cm
The magnifying power of the microscope is calculated as:
m = v0/|u0| (1 + d /fe)
= 7.2/0.9 (1 + 25/2.5 ) = 8 (1+ 10) = 88
Hence, the magnifying power of the microscope is 88.
Focal length of the objective lens, f1 = 2.0 cm and focal length of the eyepiece, f2 = 6.25 cm Distance between the objective lens and the eyepiece, d = 15 cm Ans (a). Least distance of distinct vision, d' = 25 cm Therefore, image distance for the eyepiece, v2 = -25 cm and let the object distance foRead more
Focal length of the objective lens, f1 = 2.0 cm and
focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
Ans (a).
Least distance of distinct vision, d’ = 25 cm
Therefore, image distance for the eyepiece, v2 = -25 cm and let the object distance for the eyepiece = u2
According to the lens formula, we have the relation:
1/v2 – 1/u2 = 1/f2
=> 1/u2 = 1/v2 – 1/f2
= 1/(-25) – 1/6.25 = (-1-4)/25 = -5/25
Therefore , u2 = -5 cm
Image distance for the objective lens, v1 = d + u2 = 15 — 5 = 10 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
1/v1 – 1/u1 = 1/f1
=> 1/u1 = 1/v1 – 1/f1
= 1/10 -1/2 = (1-5)/10 = -4/10
Therefore, u1 = -2.5 cm
Magnitude of the object distance, |u1| = 2.5 cm
The magnifying power of a compound microscope is given by the relation:
m = v1/|u1| x (1 + d’/f2)
= 10/2.5 (1 + 25/6.25)
= 4 (1 + 4 ) = 20
Hence, the magnifying power of the microscope is 20.
Ans (b).
The final image is formed at infinity.
Therefore ,image distance for the eyepiece, v2 =∝ and let the object distance for the eyepiece = u2
According to the lens formula, we have the relation:
1/v2 -1/u2 = 1/f2
1/∝ – 1/u2 = 1/6.25
Therefore , u2 = -6.25 cm
Image distance for the objective lens, v1 = d + u1 = 15 — 6.25 = 8.75 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
Focal length of the convex lens, f1 = 30 cm and 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 Therefore , f = -60 cm HencRead more
Focal length of the convex lens, f1 = 30 cm and
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
Therefore ,
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.
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 lens formula, we have the relation: 1/v -1/u = 1/f => 1/v -1/21 -1/14 = (-2-3)/42 = -5/42 Therefore , v = -42/5 = -8.4 cm Hence, the image is formed on the oRead more
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 lens formula, we have the relation:
1/v -1/u = 1/f
=> 1/v -1/21 -1/14 = (-2-3)/42 = -5/42
Therefore ,
v = -42/5 = -8.4 cm
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
The magnification of the image is given as:
m = Image height (h2) /Object height (h1) = v/u
Therefore , h2 = (-8.4 /-14 )x 3 = 1.8 cm
Hence, the height of the image is 1.8 cm.
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance
In the given situation, the object is virtual and the image formed is real. Object distance, u = +12 cm Ans (a). Focal length of the convex lens, f = 20 cm Image distance = v According to the lens formula, we have the relation: 1/v - 1/u = 1/f 1/v -1/12 = 1/20 1/v = 1/20 + 1/12 = (3+5)/60 = 8/60 TheRead more
In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12 cm
Ans (a).
Focal length of the convex lens, f = 20 cm
Image distance = v
According to the lens formula, we have the relation:
1/v – 1/u = 1/f
1/v -1/12 = 1/20
1/v = 1/20 + 1/12 = (3+5)/60 = 8/60
Therefore , v = 60/8 = 7.5 cm
Hence, the image is formed 7.5 cm away from the lens, toward its right.
Ans (b).
Focal length of the concave lens, f = -16 cm Image distance = v
According to the lens formula, we have the relation:
1/v – 1/u = 1/f
1/v – 1/12 = 1/16
=> 1/v = 1/(-16) + 1/12 = -1/16 + 1/12
=>1/v = (-3+ 4)/48 =1/48
v = 48 cm
Hence, the image is formed 48 cm away from the lens, toward its right.
Refractive index of glass, μ = 1.55 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 the lens = R2 Radius of curvature of the double-convex lens = R Therefore, R1= R and R2 = -R The value of R can be calculateRead more
Refractive index of glass, μ = 1.55
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 the lens = R2
Radius of curvature of the double-convex lens = R
Therefore,
R1= R and R2 = -R
The value of R can be calculated as:
1/f = (μ-1) [1/R1 -1/R2]
1/20 = (1.55 -1)[1/R + 1/R]
1/20 = 0.55 x 2/R
Therefore,
R = 0.55 x 2 x 20 = 22cm
Hence, the radius of curvature of the double-convex lens is 22 cm.
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Actual depth of the pin, d = 15 cm Apparent depth of the pin = d' Refractive index of glass,μ = 1.5 Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e. μ = d/d' Therefore ,d' = d/μ = 15/1.5 = 10 cm The distance at which the pin appears to be raised = d' - d =Read more
Actual depth of the pin, d = 15 cm
Apparent depth of the pin = d’
Refractive index of glass,μ = 1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
μ = d/d’
Therefore ,d’ = d/μ
= 15/1.5 = 10 cm
The distance at which the pin appears to be raised = d’ – d = 15 – 10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.
See lessUse the mirror equation to deduce that: (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
(a) For a concave mirror, the focal length (f) is negative. Therefore ,f < 0 When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore u < 0 For image distance v, we can write the lens formula as: 1/v - 1/u = 1/f 1/v = 1/f - 1/u------------------Read more
(a) For a concave mirror, the focal length (f) is negative. Therefore ,f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore u < 0 For image distance v, we can write the lens formula as:
1/v – 1/u = 1/f
1/v = 1/f – 1/u——————- Eq-1
The object lies between f and 2f.
However , 2f <u <f (because u and f are negative)
1/2f > 1/u > 1/f
-1/2f < -1/u <-1/f
1/f -1/2f <1/f -1/u<0 —————-Eq-2
Using equation (1), we get:
1/2f < 1/v < 0
Therefore, 1/v is negative, i.e., v is negative.
1/2f < 1/v => 2F > v
=> -v > -2f ,therefore, the image lies beyond 2f.
Ans (b).
For a convex mirror, the focal length (f) is positive. Therefore f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore, u < 0 For image distance v, we have the mirror formula:
1/v +1/u = 1/f
1/v = 1/f -1/u
Using equation (2), we can conclude that:
1/v<o
0r v>0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
Ans (c).
For a convex mirror, the focal length (f) is positive. f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative, therefore, u < 0 For image distance v, we have the mirror formula:
1/v + 1/u = 1/f
=> 1/v = 1/f -1/u
But we have u < 0
Therefore , 1/v > 1/f
or v <f
Hence, the image formed is diminished and is located between the focus (f) and the pole.
Ans (d).
For a concave mirror, the focal length (f) is negative. Therefore, f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative. Therefore, u < 0 It is placed between the focus (f) and the pole.
Therefore ,f > u> 0
1/f < 1/u < 0
1/f -1/u < 0
For image distance v, we have the mirror formula:
1/v + 1/u = 1/f
1/v = 1/f -1/u
Therefore ,1/v < 0
v >0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write: 1/u > 1/v => v > u
Magnification, m = v/u > 1 Hence, the formed image is enlarged.
See less(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 x 106 m, and the radius of lunar orbit is 3.8 x 108 m.
Focal length of the objective lens, fo = 15 m = 15 x 10² cm and focal length of the eyepiece, fe = 1.0 cm Ans (a). The angular magnification of a telescope is given as: α = fo/fe = 15 x 10²/1.0 = 1500 Hence, the angular magnification of the given refracting telescope is 1500. Ans (b). Diameter of tRead more
Focal length of the objective lens, fo = 15 m = 15 x 10² cm and focal length of the eyepiece, fe = 1.0 cm
Ans (a).
The angular magnification of a telescope is given as:
α = fo/fe = 15 x 10²/1.0 = 1500
Hence, the angular magnification of the given refracting telescope is 1500.
Ans (b).
Diameter of the moon, d = 3.48 x 10⁶ m and radius of the lunar orbit, ro = 3.8 x 10⁸ m Let d’ be the diameter of the image of the moon formed by the objective lens.
The angle subtended by the diameter of the moon is equal to the angle subtended by the image.
d/r0 = d’/f0
(3.48 x 10⁶)/(3.8 x 10⁸) = d’/15
d’= (3.48/3.8) x 10⁻² x 15
= 13.74 x 10⁻² m = 13.74 cm
Hence, the diameter of the moon’s image formed by the objective lens is 13.74 cm
See lessA small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Focal length of the objective lens, f0 = 144 cm Focal length of the eyepiece, fe = 6.0 cm The magnifying power of the telescope is given as: m = f0 /fe = 144/6 = 24 The separation between the objective lens and the eyepiece is calculated as: f0 + fe = 144 + 6 = 150 cm Hence, the magnifying poweRead more
Focal length of the objective lens, f0 = 144 cm
Focal length of the eyepiece, fe = 6.0 cm
The magnifying power of the telescope is given as:
m = f0 /fe = 144/6 = 24
The separation between the objective lens and the eyepiece is calculated as:
f0 + fe = 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.
See lessA person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Focal length of the objective lens, f0 = 8 mm = 0.8 cm Focal length of the eyepiece, fe = 2.5 cm Object distance for the objective lens, u0 = -9.0 mm = -0.9 cm Least distance of distant vision, d = 25 cm Image distance for the eyepiece, ve = -d = -25 cm Object distance for the eyepiece = ue Using thRead more
Focal length of the objective lens, f0 = 8 mm = 0.8 cm
Focal length of the eyepiece, fe = 2.5 cm
Object distance for the objective lens, u0 = -9.0 mm = -0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, ve = -d = -25 cm
Object distance for the eyepiece = ue
Using the lens formula, we can obtain the value of ue as:
1/ve – 1/ue = 1/fe
1/ue = 1/fe – 1/ve
= 1/-25 – 1/2.5 = (-1-10)/25 = -11/25
Therefor, ue = – 25/11 = – 2.27
We can also obtain the value of the image distance for the objective lens ( v0) using the lens formula.
1/v0 -1/u0 = 1/f0
1/v0 = 1/f0+ 1/u0
= 1/0.8 -1/0.9 = (0.9 -0.8)/ 0.72 = 0.1/0.72
Therefore, v0 = 7.2 cm
The distance between the objective lens and the eyepiece
= |ue| + v0 = 2.27 + 7.2 = 9.47 cm
The magnifying power of the microscope is calculated as:
m = v0/|u0| (1 + d /fe)
= 7.2/0.9 (1 + 25/2.5 ) = 8 (1+ 10) = 88
Hence, the magnifying power of the microscope is 88.
See lessA compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Focal length of the objective lens, f1 = 2.0 cm and focal length of the eyepiece, f2 = 6.25 cm Distance between the objective lens and the eyepiece, d = 15 cm Ans (a). Least distance of distinct vision, d' = 25 cm Therefore, image distance for the eyepiece, v2 = -25 cm and let the object distance foRead more
Focal length of the objective lens, f1 = 2.0 cm and
focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
Ans (a).
Least distance of distinct vision, d’ = 25 cm
Therefore, image distance for the eyepiece, v2 = -25 cm and let the object distance for the eyepiece = u2
According to the lens formula, we have the relation:
1/v2 – 1/u2 = 1/f2
=> 1/u2 = 1/v2 – 1/f2
= 1/(-25) – 1/6.25 = (-1-4)/25 = -5/25
Therefore , u2 = -5 cm
Image distance for the objective lens, v1 = d + u2 = 15 — 5 = 10 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
1/v1 – 1/u1 = 1/f1
=> 1/u1 = 1/v1 – 1/f1
= 1/10 -1/2 = (1-5)/10 = -4/10
Therefore, u1 = -2.5 cm
Magnitude of the object distance, |u1| = 2.5 cm
The magnifying power of a compound microscope is given by the relation:
m = v1/|u1| x (1 + d’/f2)
= 10/2.5 (1 + 25/6.25)
= 4 (1 + 4 ) = 20
Hence, the magnifying power of the microscope is 20.
Ans (b).
The final image is formed at infinity.
Therefore ,image distance for the eyepiece, v2 =∝ and let the object distance for the eyepiece = u2
According to the lens formula, we have the relation:
1/v2 -1/u2 = 1/f2
1/∝ – 1/u2 = 1/6.25
Therefore , u2 = -6.25 cm
Image distance for the objective lens, v1 = d + u1 = 15 — 6.25 = 8.75 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
1/v1 – 1/u1 = 1/f1
1/u1 = 1/v1 – 1/f1 = 1/8.75 – 1/2.0 = (2-8.75)/17.5
Therefore ,u1 = -17.5/6.75 = – 2.59 cm
Magnitude of the object distance, |u1| = 2.59 cm
The magnifying power of a compound microscope is given by the relation:
m = v1/|u1| x (d’/|u2|) = (8.75 /2.59) x (25 /6.25) = 13.51
Hence, the magnifying power of the microscope is 13.51.
See lessWhat 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.
Focal length of the convex lens, f1 = 30 cm and 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 Therefore , f = -60 cm HencRead more
Focal length of the convex lens, f1 = 30 cm and
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
Therefore ,
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.
See lessAn object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
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 lens formula, we have the relation: 1/v -1/u = 1/f => 1/v -1/21 -1/14 = (-2-3)/42 = -5/42 Therefore , v = -42/5 = -8.4 cm Hence, the image is formed on the oRead more
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 lens formula, we have the relation:
1/v -1/u = 1/f
=> 1/v -1/21 -1/14 = (-2-3)/42 = -5/42
Therefore ,
v = -42/5 = -8.4 cm
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
The magnification of the image is given as:
m = Image height (h2) /Object height (h1) = v/u
Therefore , h2 = (-8.4 /-14 )x 3 = 1.8 cm
Hence, the height of the image is 1.8 cm.
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance
See lessA beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
In the given situation, the object is virtual and the image formed is real. Object distance, u = +12 cm Ans (a). Focal length of the convex lens, f = 20 cm Image distance = v According to the lens formula, we have the relation: 1/v - 1/u = 1/f 1/v -1/12 = 1/20 1/v = 1/20 + 1/12 = (3+5)/60 = 8/60 TheRead more
In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12 cm
Ans (a).
Focal length of the convex lens, f = 20 cm
Image distance = v
According to the lens formula, we have the relation:
1/v – 1/u = 1/f
1/v -1/12 = 1/20
1/v = 1/20 + 1/12 = (3+5)/60 = 8/60
Therefore , v = 60/8 = 7.5 cm
Hence, the image is formed 7.5 cm away from the lens, toward its right.
Ans (b).
Focal length of the concave lens, f = -16 cm Image distance = v
According to the lens formula, we have the relation:
1/v – 1/u = 1/f
1/v – 1/12 = 1/16
=> 1/v = 1/(-16) + 1/12 = -1/16 + 1/12
=>1/v = (-3+ 4)/48 =1/48
v = 48 cm
Hence, the image is formed 48 cm away from the lens, toward its right.
See lessDouble-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Refractive index of glass, μ = 1.55 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 the lens = R2 Radius of curvature of the double-convex lens = R Therefore, R1= R and R2 = -R The value of R can be calculateRead more
Refractive index of glass, μ = 1.55
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 the lens = R2
Radius of curvature of the double-convex lens = R
Therefore,
R1= R and R2 = -R
The value of R can be calculated as:
1/f = (μ-1) [1/R1 -1/R2]
1/20 = (1.55 -1)[1/R + 1/R]
1/20 = 0.55 x 2/R
Therefore,
R = 0.55 x 2 x 20 = 22cm
Hence, the radius of curvature of the double-convex lens is 22 cm.
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