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For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m?
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Solution:
Given, magnification (m) when object distance (u) = 0.15 m is twice the magnification when u = 0.20 m.
Let the focal length of the lens be f.
Using the lens formula, we have:
1/f = 1/u + 1/v
where v is the image distance.
Using the magnification formula, we have:
m = v/u
When the object is at a distance u = 0.15 m, the magnification is given as:
m1 = v/u = 2
Substituting u = 0.15 m in the lens formula and solving for v, we get:
1/f = 1/u + 1/v
1/f = 1/0.15 + 1/v
v = 0.075 m
When the object is at a distance u = 0.20 m, the magnification is given as:
m2 = v/u
Substituting u = 0.20 m in the lens formula and solving for v, we get:
1/f = 1/u + 1/v
1/f = 1/0.20 + 1/v
v = 0.10 m
We know that m1 = 2*m2
Therefore, 2 = v1/u1 = 2*v2/u2
2 = 0.075/0.15 * u2/0.20
u2 = 0.10 m
Therefore, the object distance when the magnification is twice as large is u2 = 0.10 m.
Now, substituting u2 = 0.10 m in the lens formula and solving for f, we get:
1/f = 1/0.10 + 1/0.10
f = 0.05 m
Hence, the focal length of the lens is 0.05 m.
Answer: (4) 0.05m
Given, magnification (m) when object distance (u) = 0.15 m is twice the magnification when u = 0.20 m.
Let the focal length of the lens be f.
Using the lens formula, we have:
1/f = 1/u + 1/v
where v is the image distance.
Using the magnification formula, we have:
m = v/u
When the object is at a distance u = 0.15 m, the magnification is given as:
m1 = v/u = 2
Substituting u = 0.15 m in the lens formula and solving for v, we get:
1/f = 1/u + 1/v
1/f = 1/0.15 + 1/v
v = 0.075 m
When the object is at a distance u = 0.20 m, the magnification is given as:
m2 = v/u
Substituting u = 0.20 m in the lens formula and solving for v, we get:
1/f = 1/u + 1/v
1/f = 1/0.20 + 1/v
v = 0.10 m
We know that m1 = 2*m2
Therefore, 2 = v1/u1 = 2*v2/u2
2 = 0.075/0.15 * u2/0.20
u2 = 0.10 m
Therefore, the object distance when the magnification is twice as large is u2 = 0.10 m.
Now, substituting u2 = 0.10 m in the lens formula and solving for f, we get:
1/f = 1/0.10 + 1/0.10
f = 0.05 m
Hence, the focal length of the lens is 0.05 m.
Answer: (4) 0.05m
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For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m?
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For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m?.
For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a given lens, the magnification was found to be twice as large as when the object was 0.15 m from it as when the distance was 0.2m. The focal length of the lens is (1)1.5m. (2)0.20m (3)0.10m (4)0.05m?.
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