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A plano convex lens of refractive index 1.5 and radius of curvature 30 cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object
- a)60 cm
- b)30 cm
- c)20 cm
- d)80 cm
Correct answer is option 'C'. Can you explain this answer?
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A plano convex lens of refractive index 1.5 and radius of curvature 30...
KEY CONCEPT : The focal length(F) of the final mirror
The combination acts as a converging mirror. For the object to be of the same size of mirror, u = 2F = 20 cm
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A plano convex lens of refractive index 1.5 and radius of curvature 30...
To determine the distance at which an object must be placed in order to form a real image of the same size using a plano-convex lens that has been silvered at the curved surface, we can use the lens formula and magnification formula.
Given data:
Refractive index (n) = 1.5
Radius of curvature (R) = 30 cm
1. Lens formula:
The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens. It is given by:
1/f = 1/v - 1/u
In this case, since the lens is plano-convex, the focal length (f) can be calculated using the following formula:
f = R/2
Substituting the given values, we find:
f = 30/2 = 15 cm
2. Magnification formula:
The magnification (m) of a lens is given by the ratio of the height of the image (h') to the height of the object (h). It is given by:
m = -v/u
In this case, since we want the image to be of the same size as the object, the magnification will be 1. Therefore:
m = 1
3. Calculating the object distance:
Using the magnification formula, we can write:
m = -v/u
Since m = 1, we have:
1 = -v/u
Simplifying, we find:
u = -v
This means the object distance (u) is equal to the negative of the image distance (v).
4. Applying the lens formula:
Substituting the value of f and the relationship between u and v into the lens formula, we have:
1/f = 1/v - 1/u
1/15 = 1/v - 1/-v
Simplifying, we find:
1/15 = (2v + v)/(v^2)
15v^2 = 3v + v^2
14v^2 - 3v = 0
Solving this quadratic equation, we find two possible values for v: v = 0 or v = 3/14 cm.
Since a real image is formed when v < 0,="" we="" discard="" the="" positive="" />
5. Calculating the object distance:
Since u = -v, we find:
u = -(-3/14) = 3/14 cm
Converting to centimeters, we have:
u = 3/14 * 100 = 21.43 cm
Therefore, the object must be placed at a distance of approximately 21.43 cm from the lens in order to form a real image of the same size.
Given data:
Refractive index (n) = 1.5
Radius of curvature (R) = 30 cm
1. Lens formula:
The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens. It is given by:
1/f = 1/v - 1/u
In this case, since the lens is plano-convex, the focal length (f) can be calculated using the following formula:
f = R/2
Substituting the given values, we find:
f = 30/2 = 15 cm
2. Magnification formula:
The magnification (m) of a lens is given by the ratio of the height of the image (h') to the height of the object (h). It is given by:
m = -v/u
In this case, since we want the image to be of the same size as the object, the magnification will be 1. Therefore:
m = 1
3. Calculating the object distance:
Using the magnification formula, we can write:
m = -v/u
Since m = 1, we have:
1 = -v/u
Simplifying, we find:
u = -v
This means the object distance (u) is equal to the negative of the image distance (v).
4. Applying the lens formula:
Substituting the value of f and the relationship between u and v into the lens formula, we have:
1/f = 1/v - 1/u
1/15 = 1/v - 1/-v
Simplifying, we find:
1/15 = (2v + v)/(v^2)
15v^2 = 3v + v^2
14v^2 - 3v = 0
Solving this quadratic equation, we find two possible values for v: v = 0 or v = 3/14 cm.
Since a real image is formed when v < 0,="" we="" discard="" the="" positive="" />
5. Calculating the object distance:
Since u = -v, we find:
u = -(-3/14) = 3/14 cm
Converting to centimeters, we have:
u = 3/14 * 100 = 21.43 cm
Therefore, the object must be placed at a distance of approximately 21.43 cm from the lens in order to form a real image of the same size.
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A plano convex lens of refractive index 1.5 and radius of curvature 30 cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the objecta)60 cmb)30 cmc)20 cmd)80 cmCorrect answer is option 'C'. Can you explain this answer?
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A plano convex lens of refractive index 1.5 and radius of curvature 30 cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the objecta)60 cmb)30 cmc)20 cmd)80 cmCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A plano convex lens of refractive index 1.5 and radius of curvature 30 cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the objecta)60 cmb)30 cmc)20 cmd)80 cmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A plano convex lens of refractive index 1.5 and radius of curvature 30 cm. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the objecta)60 cmb)30 cmc)20 cmd)80 cmCorrect answer is option 'C'. Can you explain this answer?.
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