Problem 2880 - JEE Physics

Question

A convex lens of focal length \(\mathrm{f}\) is placed somewhere in between an object and a screen. The distance between the object and the screen is \(\mathrm{x}\). If the numerical value of the magnification product by the lens is \(\mathrm{m}\), What is the focal length of the lens? (A) \(\left[\mathrm{mx} /(\mathrm{m}-1)^{2}\right]\) (B) \(\left[\mathrm{mx} /(\mathrm{m}+1)^{2}\right]\) (C) \(\left[(m-1)^{2} / \mathrm{m}\right] \mathrm{x}\) (D) \(\left[(\mathrm{m}+1)^{2} / \mathrm{m}\right] \mathrm{x}\)

Step-by-Step Solution

Answer

The short answer is: The focal length of the lens is given by (D) \(\left[(\mathrm{m}+1)^{2} / \mathrm{m}\right] \mathrm{x}\).

Step 1: Write down the lens formula

The lens formula is given by \(\frac{1}{\mathrm{f}} = \frac{1}{\mathrm{v}} - \frac{1}{\mathrm{u}}\), where f is the focal length of the lens and v, u are the image and object distances from the lens respectively.

Step 2: Write down the magnification formula

The magnification formula for lenses is given by \(\mathrm{m} = - \frac{\mathrm{v}}{\mathrm{u}}\). This indicates that the ratio of the image distance to the object distance will be equal to the magnification of the lens.

Step 3: Substitute the distance between object and screen into the formula

It is given that the distance between object and screen (x) is equal to the sum of the object and image distances i.e. x = u + v.

Step 4: Express v in terms of u and m from the magnification formula

Using the magnification formula, we have v = -mu. Substitute this in x = u + v. So, x = u - mu.

Step 5: Calculate the value of u in terms of m and x

Using the above equation, we can calculate u as follows: u = x / (m - 1).

Step 6: Calculate the value of v in terms of m and x

Now substitute the value of u from the previous step into the magnification formula v = -mu to find the value of v. v = -mx/ (m - 1).

Step 7: Substitute the values of u and v in the lens formula

Now substitute these values of u and v in terms of m and x into the lens formula: \(\frac{1}{\mathrm{f}} = \frac{1}{- \mathrm{mx} / (\mathrm{m} - 1)} - \frac{1}{\mathrm{x} / (\mathrm{m} - 1)}\).

Step 8: Solve for f

Now, solve for f: f = \(\frac{(\mathrm{m} - 1)^2}{\mathrm{m}} \cdot \mathrm{x}\). Hence, the correct option is (D) \(\left[(\mathrm{m}+1)^{2} / \mathrm{m}\right] \mathrm{x}\).

Key Concepts

Convex Lens

A convex lens is a type of lens that bulges outward in the middle and is thinner at the edges. It is also known as a converging lens because it bends light rays inward. When parallel rays of light pass through a convex lens, they converge at a point on the other side known as the focal point. This property makes convex lenses useful in applications like magnifying glasses, cameras, and eyeglasses for farsightedness.
In practical situations, a convex lens can be used to produce both real and virtual images depending on the position of the object relative to the lens. The focal point and the focal length of the lens are crucial in determining the behavior and characteristics of the images formed by a convex lens.

Focal Length

The focal length of a lens is the distance between the center of the lens and its focal point. It is denoted as \(f\) and is measured in units like millimeters or meters.
The focal length determines how strongly the lens converges or diverges light. For a convex lens, the focal length is positive and influences how large or small an image is, and whether it will be real or virtual. In calculations, the lens formula \( \frac{1}{\mathrm{f}} = \frac{1}{\mathrm{v}} - \frac{1}{\mathrm{u}} \) is used to relate focal length (f) with image distance (v) and object distance (u).
Understanding focal length is vital for designing optical devices, as it affects magnification and image formation. Shorter focal lengths bend light more sharply than longer ones, influencing the field of view and image size.

Magnification

Magnification defines how much larger or smaller an image is compared to the object. It is given by the magnification formula \( \mathrm{m} = - \frac{\mathrm{v}}{\mathrm{u}} \).
A positive magnification value indicates that the image is upright relative to the object, while a negative value shows it is inverted. Magnification is an essential characteristic of optical systems as it determines how images appear to the viewer.
In our exercise, magnification plays a key role in finding the focal length, with the provided magnification affecting the relationship between the object distance, image distance, and the overall system behavior. The exercise showcases how changes in these relationships alter the focal properties of the lens used.

Object and Image Distance

Object distance (u) and image distance (v) are fundamental concepts in optics. These distances measure how far an object and its corresponding image are from the lens.
The object distance is denoted as \(u\) and is the distance from the object to the optical center of the lens. Likewise, the image distance is represented as \(v\), describing how far the image is from the lens.
These distances work together with the lens formula to determine how and where an image will be formed by the lens, whether it is real or virtual, and its orientation (inverted or upright). By manipulating object and image distances, we control image size and clarity, which is critical in devices such as cameras, microscopes, and eyewear.