Angle of deflection, 0 = 3.5° Distance of the screen from the mirror, D = 1.5 m The reflected rays get deflected by an amount twice the angle of deflection i.e., 20 = 7.0° The displacement (d) of the reflected spot of light on the screen is given as: tan 20 = d/1.5 => d = 1.5 x tan 7° = 0.184 m =Read more
Angle of deflection, 0 = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 20 = 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
tan 20 = d/1.5 => d = 1.5 x tan 7° = 0.184 m = 18.4 cm
Hence, the displacement of the reflected spot of light is 18.4 cm.
The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror. Distance between the objective mirror and the secondary mirror, d = 20 mm Radius of curvature of the objective mirror, R1= 220 mm Hence, focal length of the objective mirror, f1= R1/R2 = 110 mm RadiRead more
The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.
Distance between the objective mirror and the secondary mirror, d = 20 mm
Radius of curvature of the objective mirror, R1= 220 mm
Hence, focal length of the objective mirror, f1= R1/R2 = 110 mm
Radius of curvature of the secondary mirror, R2 = 140 mm
Hence, focal length of the secondary mirror, f2 =R2 /2= 70 mm
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror, u = f1-d = 110 -20 =90mm
Applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:
1/v + 1/u = 1/f2
1/v = 1/f2– 1/u
= 1/70 -1/90 = (9-7)/630 = 2/630
Therefore ,V = 630/2 = 315 mm
Hence, the final image will be formed 315 mm away from the secondary mirror.
Focal length of the objective lens, f0 = 140 cm Focal length of the eyepiece, fe = 5 cm Ans (a). In normal adjustment, the separation between the objective lens and the eyepiece = f0 + fe = 140 + 5 = 145 cm Ans (b). Height of the tower, h₁= 100 m Distance of the tower (object) from the telescope, uRead more
Focal length of the objective lens, f0 = 140 cm
Focal length of the eyepiece, fe = 5 cm
Ans (a).
In normal adjustment, the separation between the objective lens and the eyepiece = f0 + fe = 140 + 5 = 145 cm
Ans (b).
Height of the tower, h₁= 100 m
Distance of the tower (object) from the telescope, u = 3 km = 3000 m The angle subtended by the tower at the telescope is given as:
0 = h1/u = 100/3000 = 1/30 rad
The angle subtended by the image produced by the objective lens is given as:
0 = h2/f0= h2/140 rad
Where, h2 = Height of the image of the tower formed by the objective lens
1/30 = h2/140
Therefore , h2 = 140/30 = 4.7 cm
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
Ans (c).
Image is formed at a distance, d = 25 cm
The magnification of the eyepiece is given by the relation:
m = 1 + d/fe
= 1+ 25/5 = 1 + 5 =6
Height of the final image = mh2 = 6 x 4.7 = 28.2 cm
Hence, the height of the final image of the tower is 28.2 cm
Focal length of the objective lens, f0 = 140 cm Focal length of the eyepiece, fe = 5 cm Least distance of distinct vision, d = 25 cm Ans (a). When the telescope is in normal adjustment, its magnifying power is given as: m = f0/fe =140/5 = 28 Ans (b). When the final image is formed at d, the magnifyiRead more
Focal length of the objective lens, f0 = 140 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Ans (a).
When the telescope is in normal adjustment, its magnifying power is given as:
m = f0/fe =140/5 = 28
Ans (b).
When the final image is formed at d, the magnifying power of the telescope is given as:
Focal length of the objective lens, f0 = 1.25 cm Focal length of the eyepiece, fe = 5 cm Least distance of distinct vision, d = 25 cm Angular magnification of the compound microscope = 30X Total magnifying power of the compound microscope, m = 30 The angular magnification of the eyepiece is given byRead more
Focal length of the objective lens, f0 = 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
me = ( 1 + d/fe) = (1 + 25/5 ) = 6
The angular magnification of the objective lens (m0) is related to me as: m0me = m
m0= m/me = 30/6 = 5
We also have the relation:
m0 = Image distance for the objective lens (v0)/Object distance for the objective lens (-u0)
5 = v0/(-u0)
Therefore , v0= -5u0————–Eq-1
Applying the lens formula for the objective lens:
1/f0 =1/v0 -1/u 0
1/1.25 = 1/(-5u0) – 1/u0) = -6/5u0
Therefore u0 = -6/5 x 1.25 = -1.5 cm
and v0 = -5u0
= -5 x (-1.5) = 7.5cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
1/ve -1/ue = 1/fe
Where, ve = Image distance for the eyepiece = – d = – 25 cm
ue = Object distance for the eyepiece
1/ue = 1/ve -1/fe
= -1/25 -1/5 = -6/25
Therefore , ue =-4.17 cm
Separation between the objective lens and the eyepiece = |ue| + |v0| = 4.17 + 7.5 = 11.67 cm Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
Ans (a). Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magRead more
Ans (a).
Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
Ans (b).
Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
Ans (c).
The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.
Ans (d).
The angular magnification produced by the eyepiece of a compound microscope is [(25/fe) + 1 }
Where, fe = Focal length of the eyepiece
It can be inferred that if fe is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as 1/(|u0/f0|)
Where, u0 = Object distance for the objective lens and f0 = Focal length of the objective
The magnification is large when u0> f0. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since U0 is small, f0 will be even smaller. Therefore, fe and f0 are both small in the given condition.
Ans (e).
When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.
Area of the virtual image of each square, A = 6.25 mm2 Area of each square, Ao = 1 mm2 Hence, the linear magnification of the object can be calculated as: m = √ (A/A0 ) = √ (6.25/1 ) =2.5 But ,m = Image distance (v) /Object distance (u) Therefore v= mu = 2.5u Focal length of the amgnifying glass ,fRead more
Area of the virtual image of each square, A = 6.25 mm2
Area of each square, Ao = 1 mm2
Hence, the linear magnification of the object can be calculated as:
m = √ (A/A0 ) = √ (6.25/1 ) =2.5
But ,m = Image distance (v) /Object distance (u)
Therefore v= mu = 2.5u
Focal length of the amgnifying glass ,f = 10 cm.According to the lens formula,we have the relation:
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
Ans (a). The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm). Image distance, v = -d = -25 cm Focal length, f = 10 cm Object distance = u According to the lens formula, we have: 1/f =1/v -1/u 1/u = 1/v-1/f = 1/(-25) -1/10 = (-2-5)/50 = -7/50 ThereforRead more
Ans (a).
The maximum possible magnification is obtained when the image is formed
at the near point (d = 25 cm). Image distance,
v = -d = -25 cm
Focal length, f = 10 cm
Object distance = u
According to the lens formula, we have:
1/f =1/v -1/u
1/u = 1/v-1/f
= 1/(-25) -1/10 = (-2-5)/50 = -7/50
Therefore, u = -50/7 = -7.14 cm
Hence, to view the squares distinctly ,the lens should be kept 7.14 cm away from them.
Ans (b).
Magnification = |v/u| = 25 /(50/7) = 3.5
Ans (c).
Magnifying Power = d/u = 25 /(50/7) = 3.5
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
Ans (a). Area of each square, A = 1 mm2 Object distance, u = -9 cm Focal length of a converging lens, f = 10 cm For image distance v, the lens formula can be written as: 1/f = 1/v -1/u 1/10 = 1/v + 1/9 1/v = -1/90 Therefore v= -90 cm Magnification ,m= v/u = -90/(-9) =10 Therefore ,area of each squarRead more
Ans (a).
Area of each square, A = 1 mm2
Object distance, u = -9 cm
Focal length of a converging lens, f = 10 cm
For image distance v, the lens formula can be written as:
1/f = 1/v -1/u
1/10 = 1/v + 1/9
1/v = -1/90
Therefore v= -90 cm
Magnification ,m= v/u
= -90/(-9) =10
Therefore ,area of each square in the virtual image = (10)2 A = 102 x 1 = 100 mm2 = 1 cm2
Ans (b).
Magnifying power of the lens = d/|u| = 25/9 = 2.8
Ans (c).
The magnification in (a) is not the same as the magnifying power in (b).
The magnification magnitude is |v/u| and the magnifying power is (d/|u|)
The two quantities will be equal when the image is formed at the near point (25 cm).
Ans (a). Focal length of the magnifying glass, f = 5 cm Least distance of distance vision, d = 25 cm, closest object distance = u and image distance, v = -d = -25 cm According to the lens formula, we have: 1/f = 1/v - 1/u => 1/u = 1/v -1/f = 1/(-25) - 1/5 = (-5-1)/25 = -6/25 Therefore u = -25/6 =Read more
Ans (a).
Focal length of the magnifying glass, f = 5 cm
Least distance of distance vision, d = 25 cm, closest object distance = u and image distance,
v = -d = -25 cm
According to the lens formula, we have:
1/f = 1/v – 1/u
=> 1/u = 1/v -1/f
= 1/(-25) – 1/5 = (-5-1)/25 = -6/25
Therefore u = -25/6 = -4.167 cm
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distant (u’), the image distance (v’) = ∝ .
According to the lens formula, we have:
1/f = 1/v’ -1/u’
1/u’ =1/∝ -1/5 =-1/5
Therefore u’ = -5cm
Hence, the farthest distance at which the person can read the book is 5 cm.
Ans (b).
Maximum angular magnification is given by the relation:
αmax
= d/|u| = 25/(25/6) =6
Minimum angular magnification is given by the relation:
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Figure. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
Angle of deflection, 0 = 3.5° Distance of the screen from the mirror, D = 1.5 m The reflected rays get deflected by an amount twice the angle of deflection i.e., 20 = 7.0° The displacement (d) of the reflected spot of light on the screen is given as: tan 20 = d/1.5 => d = 1.5 x tan 7° = 0.184 m =Read more
Angle of deflection, 0 = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 20 = 7.0°
The displacement (d) of the reflected spot of light on the screen is given as:
tan 20 = d/1.5 => d = 1.5 x tan 7° = 0.184 m = 18.4 cm
Hence, the displacement of the reflected spot of light is 18.4 cm.
See lessA Cassegrain telescope uses two mirrors as shown in Figure.9.33 Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror. Distance between the objective mirror and the secondary mirror, d = 20 mm Radius of curvature of the objective mirror, R1= 220 mm Hence, focal length of the objective mirror, f1= R1/R2 = 110 mm RadiRead more
The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.
Distance between the objective mirror and the secondary mirror, d = 20 mm
Radius of curvature of the objective mirror, R1= 220 mm
Hence, focal length of the objective mirror, f1= R1/R2 = 110 mm
Radius of curvature of the secondary mirror, R2 = 140 mm
Hence, focal length of the secondary mirror, f2 =R2 /2= 70 mm
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror, u = f1-d = 110 -20 =90mm
Applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:
1/v + 1/u = 1/f2
1/v = 1/f2– 1/u
= 1/70 -1/90 = (9-7)/630 = 2/630
Therefore ,V = 630/2 = 315 mm
Hence, the final image will be formed 315 mm away from the secondary mirror.
See less(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? (c) What is the height of the final image of the tower if it is formed at 25 cm?
Focal length of the objective lens, f0 = 140 cm Focal length of the eyepiece, fe = 5 cm Ans (a). In normal adjustment, the separation between the objective lens and the eyepiece = f0 + fe = 140 + 5 = 145 cm Ans (b). Height of the tower, h₁= 100 m Distance of the tower (object) from the telescope, uRead more
Focal length of the objective lens, f0 = 140 cm
Focal length of the eyepiece, fe = 5 cm
Ans (a).
In normal adjustment, the separation between the objective lens and the eyepiece = f0 + fe = 140 + 5 = 145 cm
Ans (b).
Height of the tower, h₁= 100 m
Distance of the tower (object) from the telescope, u = 3 km = 3000 m The angle subtended by the tower at the telescope is given as:
0 = h1/u = 100/3000 = 1/30 rad
The angle subtended by the image produced by the objective lens is given as:
0 = h2/f0= h2/140 rad
Where, h2 = Height of the image of the tower formed by the objective lens
1/30 = h2/140
Therefore , h2 = 140/30 = 4.7 cm
Therefore, the objective lens forms a 4.7 cm tall image of the tower.
Ans (c).
Image is formed at a distance, d = 25 cm
The magnification of the eyepiece is given by the relation:
m = 1 + d/fe
= 1+ 25/5 = 1 + 5 =6
Height of the final image = mh2 = 6 x 4.7 = 28.2 cm
Hence, the height of the final image of the tower is 28.2 cm
See lessA small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25 cm)?
Focal length of the objective lens, f0 = 140 cm Focal length of the eyepiece, fe = 5 cm Least distance of distinct vision, d = 25 cm Ans (a). When the telescope is in normal adjustment, its magnifying power is given as: m = f0/fe =140/5 = 28 Ans (b). When the final image is formed at d, the magnifyiRead more
Focal length of the objective lens, f0 = 140 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Ans (a).
When the telescope is in normal adjustment, its magnifying power is given as:
m = f0/fe =140/5 = 28
Ans (b).
When the final image is formed at d, the magnifying power of the telescope is given as:
m = f0/fe [1 + fe/d]
= 140/5 [1 + 5/25] = 28 [1+ 0.2] =28 x 1.2 = 33.6
See lessAn angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Focal length of the objective lens, f0 = 1.25 cm Focal length of the eyepiece, fe = 5 cm Least distance of distinct vision, d = 25 cm Angular magnification of the compound microscope = 30X Total magnifying power of the compound microscope, m = 30 The angular magnification of the eyepiece is given byRead more
Focal length of the objective lens, f0 = 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
me = ( 1 + d/fe) = (1 + 25/5 ) = 6
The angular magnification of the objective lens (m0) is related to me as: m0me = m
m0= m/me = 30/6 = 5
We also have the relation:
m0 = Image distance for the objective lens (v0)/Object distance for the objective lens (-u0)
5 = v0/(-u0)
Therefore , v0= -5u0————–Eq-1
Applying the lens formula for the objective lens:
1/f0 =1/v0 -1/u 0
1/1.25 = 1/(-5u0) – 1/u0) = -6/5u0
Therefore u0 = -6/5 x 1.25 = -1.5 cm
and v0 = -5u0
= -5 x (-1.5) = 7.5cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
1/ve -1/ue = 1/fe
Where, ve = Image distance for the eyepiece = – d = – 25 cm
ue = Object distance for the eyepiece
1/ue = 1/ve -1/fe
= -1/25 -1/5 = -6/25
Therefore , ue =-4.17 cm
Separation between the objective lens and the eyepiece = |ue| + |v0| = 4.17 + 7.5 = 11.67 cm Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
See lessAnswer the following questions: (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? (e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Ans (a). Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magRead more
Ans (a).
Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.
Ans (b).
Yes, the angular magnification changes. When the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.
Ans (c).
The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.
Ans (d).
The angular magnification produced by the eyepiece of a compound microscope is [(25/fe) + 1 }
Where, fe = Focal length of the eyepiece
It can be inferred that if fe is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as 1/(|u0/f0|)
Where, u0 = Object distance for the objective lens and f0 = Focal length of the objective
The magnification is large when u0> f0. In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little. Since U0 is small, f0 will be even smaller. Therefore, fe and f0 are both small in the given condition.
Ans (e).
When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.
See lessWhat should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm²? Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Area of the virtual image of each square, A = 6.25 mm2 Area of each square, Ao = 1 mm2 Hence, the linear magnification of the object can be calculated as: m = √ (A/A0 ) = √ (6.25/1 ) =2.5 But ,m = Image distance (v) /Object distance (u) Therefore v= mu = 2.5u Focal length of the amgnifying glass ,fRead more
Area of the virtual image of each square, A = 6.25 mm2
Area of each square, Ao = 1 mm2
Hence, the linear magnification of the object can be calculated as:
m = √ (A/A0 ) = √ (6.25/1 ) =2.5
But ,m = Image distance (v) /Object distance (u)
Therefore v= mu = 2.5u
Focal length of the amgnifying glass ,f = 10 cm.According to the lens formula,we have the relation:
1/f = 1/v-1/u
1/10 =1/2.5u -1/u = 1/u (1/2.5 -1/1) = 1/u (1-2.5)/2.5
Therefore, u = 1.5 x 10 /2.5 = -6 cm
and v = 2.5 u = 2.5 x 6 = -15cm
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.
See less(a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case? (c) Is the magnification equal to the magnifying power in this case? Explain.
Ans (a). The maximum possible magnification is obtained when the image is formed at the near point (d = 25 cm). Image distance, v = -d = -25 cm Focal length, f = 10 cm Object distance = u According to the lens formula, we have: 1/f =1/v -1/u 1/u = 1/v-1/f = 1/(-25) -1/10 = (-2-5)/50 = -7/50 ThereforRead more
Ans (a).
The maximum possible magnification is obtained when the image is formed
at the near point (d = 25 cm). Image distance,
v = -d = -25 cm
Focal length, f = 10 cm
Object distance = u
According to the lens formula, we have:
1/f =1/v -1/u
1/u = 1/v-1/f
= 1/(-25) -1/10 = (-2-5)/50 = -7/50
Therefore, u = -50/7 = -7.14 cm
Hence, to view the squares distinctly ,the lens should be kept 7.14 cm away from them.
Ans (b).
Magnification = |v/u| = 25 /(50/7) = 3.5
Ans (c).
Magnifying Power = d/u = 25 /(50/7) = 3.5
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.
See lessA card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Ans (a). Area of each square, A = 1 mm2 Object distance, u = -9 cm Focal length of a converging lens, f = 10 cm For image distance v, the lens formula can be written as: 1/f = 1/v -1/u 1/10 = 1/v + 1/9 1/v = -1/90 Therefore v= -90 cm Magnification ,m= v/u = -90/(-9) =10 Therefore ,area of each squarRead more
Ans (a).
Area of each square, A = 1 mm2
Object distance, u = -9 cm
Focal length of a converging lens, f = 10 cm
For image distance v, the lens formula can be written as:
1/f = 1/v -1/u
1/10 = 1/v + 1/9
1/v = -1/90
Therefore v= -90 cm
Magnification ,m= v/u
= -90/(-9) =10
Therefore ,area of each square in the virtual image = (10)2 A = 102 x 1 = 100 mm2 = 1 cm2
Ans (b).
Magnifying power of the lens = d/|u| = 25/9 = 2.8
Ans (c).
The magnification in (a) is not the same as the magnifying power in (b).
The magnification magnitude is |v/u| and the magnifying power is (d/|u|)
The two quantities will be equal when the image is formed at the near point (25 cm).
See lessA man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm. (a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass? (b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
Ans (a). Focal length of the magnifying glass, f = 5 cm Least distance of distance vision, d = 25 cm, closest object distance = u and image distance, v = -d = -25 cm According to the lens formula, we have: 1/f = 1/v - 1/u => 1/u = 1/v -1/f = 1/(-25) - 1/5 = (-5-1)/25 = -6/25 Therefore u = -25/6 =Read more
Ans (a).
Focal length of the magnifying glass, f = 5 cm
Least distance of distance vision, d = 25 cm, closest object distance = u and image distance,
v = -d = -25 cm
According to the lens formula, we have:
1/f = 1/v – 1/u
=> 1/u = 1/v -1/f
= 1/(-25) – 1/5 = (-5-1)/25 = -6/25
Therefore u = -25/6 = -4.167 cm
Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distant (u’), the image distance (v’) = ∝ .
According to the lens formula, we have:
1/f = 1/v’ -1/u’
1/u’ =1/∝ -1/5 =-1/5
Therefore u’ = -5cm
Hence, the farthest distance at which the person can read the book is 5 cm.
Ans (b).
Maximum angular magnification is given by the relation:
αmax
= d/|u| = 25/(25/6) =6
Minimum angular magnification is given by the relation:
αmin
= d/|u’| = 25/5=5
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