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A body is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance 'd' ahead of second wall, then the required focal length will be [2002]
- a)only d/4
- b)only d/2
- c)more than d/4 but less than d/2
- d)less than d/4
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A body is located on a wall. Its image of equal size is to be obtained...
Using the lens formula
Given v = d, for equal size image |v | =|u| d
By sign convention u = –d
By sign convention u = –d
Most Upvoted Answer
A body is located on a wall. Its image of equal size is to be obtained...
Explanation:
Given:
- A body is located on a wall.
- Its image of equal size is to be obtained on a parallel wall with the help of a convex lens.
- The lens is placed at a distance d ahead of the second wall.
To find:
The required focal length of the lens.
Solution:
- In this scenario, we are dealing with a lens formula that relates the object distance (u), image distance (v), and focal length (f) of the lens: 1/f = 1/v + 1/u.
- Since the image needs to be of equal size on the parallel wall, the magnification factor should be 1.
- For a magnification of 1, the object distance (u) and image distance (v) will be equal.
- The distance between the lens and the second wall is given as d, so the object distance (u) and image distance (v) will both be d.
- Substituting the values of u and v into the lens formula, 1/f = 1/d + 1/d = 2/d.
- Therefore, f = d/2.
Therefore, the required focal length of the lens will be d/2.
Given:
- A body is located on a wall.
- Its image of equal size is to be obtained on a parallel wall with the help of a convex lens.
- The lens is placed at a distance d ahead of the second wall.
To find:
The required focal length of the lens.
Solution:
- In this scenario, we are dealing with a lens formula that relates the object distance (u), image distance (v), and focal length (f) of the lens: 1/f = 1/v + 1/u.
- Since the image needs to be of equal size on the parallel wall, the magnification factor should be 1.
- For a magnification of 1, the object distance (u) and image distance (v) will be equal.
- The distance between the lens and the second wall is given as d, so the object distance (u) and image distance (v) will both be d.
- Substituting the values of u and v into the lens formula, 1/f = 1/d + 1/d = 2/d.
- Therefore, f = d/2.
Therefore, the required focal length of the lens will be d/2.
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A body is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance 'd' ahead of second wall, then the required focal length will be [2002]a)only d/4b)only d/2c)more than d/4 but less than d/2d)less than d/4Correct answer is option 'B'. Can you explain this answer?
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