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A screen having a real image of magnification m1 formed by a convex lens is moved a distance x. The object is then moved until a new image of magnification m2 is formed on the screen. The focal length of lens is?
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A screen having a real image of magnification m1 formed by a convex le...
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Introduction
In this scenario, we have a convex lens that forms a real image with a magnification of m1 on a screen. The screen is then moved a distance x, and the object is moved until a new image with a magnification of m2 is formed on the screen. We need to determine the focal length of the lens.

Understanding the Situation
To begin with, let's understand the concept of magnification. Magnification (m) is the ratio of the height of the image (h_i) to the height of the object (h_o). It can be calculated using the formula: m = h_i / h_o.

Identifying the Known Values
In this situation, we are given the following information:
- Initial magnification (m1)
- Final magnification (m2)
- Distance traveled by the screen (x)

Applying the Lens Formula
The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of a lens. It is given by: 1/f = 1/v - 1/u.

Deriving the Lens Formula
To derive the lens formula, we consider the initial and final positions of the screen.

Initial Position
- Object distance (u1) = Infinity (as the object is at a large distance)
- Image distance (v1) = Distance between lens and screen
- Magnification (m1) = v1 / u1

Final Position
- Object distance (u2) = Distance between lens and object
- Image distance (v2) = Distance between lens and screen
- Magnification (m2) = v2 / u2

Applying the Lens Formula for Initial Position
Using the lens formula, we can write: 1/f = 1/v1 - 1/u1

Since u1 = Infinity, the equation simplifies to: 1/f = 1/v1

Applying the Lens Formula for Final Position
Using the lens formula, we can write: 1/f = 1/v2 - 1/u2

Using the Magnification Equations
We know that magnification (m) is equal to the ratio of image distance (v) to object distance (u).

For the initial position: m1 = v1 / Infinity = v1 / ∞ = 0

For the final position: m2 = v2 / u2

Relating m1, m2, and x
We are given the magnification values (m1 and m2) and the distance traveled by the screen (x). We need to relate these values to determine the focal length (f) of the lens.

Since magnification is the ratio of image distance to object distance, we can write:

m1 = v1 / Infinity = 0

m2 = v2 / u2

Since the screen has moved a distance x, the image distance has also changed. We can relate the image distance for the initial and final positions as:

v2 = v1 - x

Solving for f
To determine the focal length (f), we can substitute the values of m1, m2, v2, and u2 into
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A screen having a real image of magnification m1 formed by a convex lens is moved a distance x. The object is then moved until a new image of magnification m2 is formed on the screen. The focal length of lens is?
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A screen having a real image of magnification m1 formed by a convex lens is moved a distance x. The object is then moved until a new image of magnification m2 is formed on the screen. The focal length of lens is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A screen having a real image of magnification m1 formed by a convex lens is moved a distance x. The object is then moved until a new image of magnification m2 is formed on the screen. The focal length of lens is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A screen having a real image of magnification m1 formed by a convex lens is moved a distance x. The object is then moved until a new image of magnification m2 is formed on the screen. The focal length of lens is?.
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