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The image of the object formed by a convex spherical mirror is magnified one third in comparison with the onject, if the mirror is displaced by 40 cm along ints principal axis in the direction of light, the image is again found one third in comparidon with the object. Find the radius of curvature?
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The image of the object formed by a convex spherical mirror is magnifi...
Explanation:
Let the distance between the object and the mirror be u, and the distance between the image and the mirror be v. The magnification produced by the mirror is given by
m = v/u
The magnification of the image formed by the mirror is one-third that of the object, so we have
v/u = 1/3
When the mirror is displaced by 40 cm along its principal axis in the direction of the light, the distance between the object and the mirror becomes u' = u + 40 cm, and the distance between the image and the mirror becomes v' = v + 40 cm. The magnification produced by the mirror is still given by
m' = v'/u'
The magnification of the image formed by the mirror after displacement is again one-third that of the object, so we have
v'/u' = 1/3
Substituting the values of v/u and v'/u' in these equations, we get
v/u = 1/3
v'/(u + 40) = 1/3
Solving these equations for v and v', we get
v = u/3
v' = (u + 40)/3
Substituting these values in the mirror formula 1/v + 1/u = 1/f, we get
1/(u/3) + 1/u = 1/f
1/(u+40)/3 + 1/u+40 = 1/f
Simplifying these equations, we get
f = -120 cm
The negative sign indicates that the mirror is convex. Therefore, the radius of curvature of the mirror is given by
R = 2f = -240 cm.
Let the distance between the object and the mirror be u, and the distance between the image and the mirror be v. The magnification produced by the mirror is given by
m = v/u
The magnification of the image formed by the mirror is one-third that of the object, so we have
v/u = 1/3
When the mirror is displaced by 40 cm along its principal axis in the direction of the light, the distance between the object and the mirror becomes u' = u + 40 cm, and the distance between the image and the mirror becomes v' = v + 40 cm. The magnification produced by the mirror is still given by
m' = v'/u'
The magnification of the image formed by the mirror after displacement is again one-third that of the object, so we have
v'/u' = 1/3
Substituting the values of v/u and v'/u' in these equations, we get
v/u = 1/3
v'/(u + 40) = 1/3
Solving these equations for v and v', we get
v = u/3
v' = (u + 40)/3
Substituting these values in the mirror formula 1/v + 1/u = 1/f, we get
1/(u/3) + 1/u = 1/f
1/(u+40)/3 + 1/u+40 = 1/f
Simplifying these equations, we get
f = -120 cm
The negative sign indicates that the mirror is convex. Therefore, the radius of curvature of the mirror is given by
R = 2f = -240 cm.
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The image of the object formed by a convex spherical mirror is magnified one third in comparison with the onject, if the mirror is displaced by 40 cm along ints principal axis in the direction of light, the image is again found one third in comparidon with the object. Find the radius of curvature?
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The image of the object formed by a convex spherical mirror is magnified one third in comparison with the onject, if the mirror is displaced by 40 cm along ints principal axis in the direction of light, the image is again found one third in comparidon with the object. Find the radius of curvature? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The image of the object formed by a convex spherical mirror is magnified one third in comparison with the onject, if the mirror is displaced by 40 cm along ints principal axis in the direction of light, the image is again found one third in comparidon with the object. Find the radius of curvature? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The image of the object formed by a convex spherical mirror is magnified one third in comparison with the onject, if the mirror is displaced by 40 cm along ints principal axis in the direction of light, the image is again found one third in comparidon with the object. Find the radius of curvature?.
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