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 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.
See less