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A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.
The focal length of the lens is (in cm)
Correct answer is '10'. Can you explain this answer?
Most Upvoted Answer
A convex lens forms a real image of an object kept at 15 cm from the ...
Given:
- Object distance (u) = 15 cm
- Object displacement (Δu) = 0.1 cm
- Image displacement (Δv) = 0.4 cm
To find:
- Focal length of the convex lens (f)
Solution:
The given problem can be solved using the lens formula:
1/f = 1/v - 1/u
Step 1: Finding the initial image distance (v)
Given that the object distance (u) is 15 cm and a real image is formed on the screen, the image distance (v) can be positive and real.
Using the lens formula,
1/f = 1/v - 1/u
Substituting the values,
1/f = 1/v - 1/15
Solving for v,
1/v = 1/f + 1/15
v = 15/(1/f + 1/15)
v = 15f/(f + 15)
Step 2: Finding the new image distance (v + Δv)
When the object is moved by Δu = 0.1 cm, the image distance changes to v + Δv. So, we need to find the new image distance (v + Δv).
Using the lens formula,
1/f = 1/(v + Δv) - 1/(u + Δu)
Substituting the values,
1/f = 1/(v + Δv) - 1/(15 + 0.1)
Simplifying,
1/f = 1/(v + Δv) - 1/15.1
Solving for (v + Δv),
1/(v + Δv) = 1/f + 1/15.1
(v + Δv) = 15.1/(1/f + 1/15.1)
(v + Δv) = 15.1f/(f + 15.1)
Step 3: Finding the relationship between (v + Δv) and v
The problem states that when the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image. This means that the change in image distance (Δv) is four times the change in object distance (Δu).
Given, Δu = 0.1 cm and Δv = 0.4 cm
Δv = 4Δu
0.4 = 4(0.1)
0.4 = 0.4
Therefore, we can write (v + Δv) = 4v.
Step 4: Solving the equations
Substituting the expressions for (v + Δv) and v from Steps 2 and 3 into the equation (v + Δv) = 4v, we get:
15.1f/(f + 15.1) = 4(15f/(f + 15))
Simplifying the equation,
15.1f(f + 15) = 4(15f)(f + 15.1)
15.1f^2 + 15.1f^2 + 226.5f = 60f^2 + 60
- Object distance (u) = 15 cm
- Object displacement (Δu) = 0.1 cm
- Image displacement (Δv) = 0.4 cm
To find:
- Focal length of the convex lens (f)
Solution:
The given problem can be solved using the lens formula:
1/f = 1/v - 1/u
Step 1: Finding the initial image distance (v)
Given that the object distance (u) is 15 cm and a real image is formed on the screen, the image distance (v) can be positive and real.
Using the lens formula,
1/f = 1/v - 1/u
Substituting the values,
1/f = 1/v - 1/15
Solving for v,
1/v = 1/f + 1/15
v = 15/(1/f + 1/15)
v = 15f/(f + 15)
Step 2: Finding the new image distance (v + Δv)
When the object is moved by Δu = 0.1 cm, the image distance changes to v + Δv. So, we need to find the new image distance (v + Δv).
Using the lens formula,
1/f = 1/(v + Δv) - 1/(u + Δu)
Substituting the values,
1/f = 1/(v + Δv) - 1/(15 + 0.1)
Simplifying,
1/f = 1/(v + Δv) - 1/15.1
Solving for (v + Δv),
1/(v + Δv) = 1/f + 1/15.1
(v + Δv) = 15.1/(1/f + 1/15.1)
(v + Δv) = 15.1f/(f + 15.1)
Step 3: Finding the relationship between (v + Δv) and v
The problem states that when the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image. This means that the change in image distance (Δv) is four times the change in object distance (Δu).
Given, Δu = 0.1 cm and Δv = 0.4 cm
Δv = 4Δu
0.4 = 4(0.1)
0.4 = 0.4
Therefore, we can write (v + Δv) = 4v.
Step 4: Solving the equations
Substituting the expressions for (v + Δv) and v from Steps 2 and 3 into the equation (v + Δv) = 4v, we get:
15.1f/(f + 15.1) = 4(15f/(f + 15))
Simplifying the equation,
15.1f(f + 15) = 4(15f)(f + 15.1)
15.1f^2 + 15.1f^2 + 226.5f = 60f^2 + 60
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Community Answer
A convex lens forms a real image of an object kept at 15 cm from the ...
Using 
u = -15 m
v = mu
v = (-2) × (-15) = 30 cm
Hence,
f = 10 cm
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A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer?.
A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer?.
Solutions for A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer?, a detailed solution for A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer? has been provided alongside types of A convex lens forms a real image of an object kept at 15 cm from the lens on the screen. If the object is moved by 0.1 cm, the screen has to be moved by 0.4 cm to capture the image.The focal length of the lens is (in cm)Correct answer is '10'. Can you explain this answer? theory, EduRev gives you an
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