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.
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 less