Problem 41 - Master Resource Book in JEE Main Physics

Question

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall \(3 \mathrm{~m}\) away by means of a large convex lens. What is the maximum possible foeal length of the lens required for the purpose? (a) \(0.88 \mathrm{~m}\) (b) \(0.90 \mathrm{~m}\) (c) \(0.75 \mathrm{~m}\) (d) \(0.63 \mathrm{~m}\)

Step-by-Step Solution

Answer

0.75 m, option (c).

Step 1: Understand the problem

The exercise asks for the maximum focal length of a convex lens needed to project the image of a bulb on the opposite wall of a room. The walls are separated by 3 m.

Step 2: Adjust the lens equation

We use the lens formula \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), where \( f \) is the focal length, \( u \) is the object distance from the lens, and \( v \) is the image distance from the lens. Here, the object (bulb) and the image (on the opposite wall) are 3 m apart, so \( u + v = 3 \).

Step 3: Express v in terms of u

Since \( u + v = 3 \), we can express \( v = 3 - u \). Substitute \( v = 3 - u \) into the lens formula to get: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{(3-u)} \).

Step 4: Simplify the focal length equation

Combine the fractions on the right-hand side: \( \frac{1}{f} = \frac{3}{u(3-u)} \). This simplifies the expression for \( f \) as \( f = \frac{u(3-u)}{3} \).

Step 5: Determine maximum focal length

The expression \( u(3-u) \) is maximum when \( u = \frac{3}{2} \) according to the symmetry of the quadratic function \( u(3-u) = 3u - u^2 \). At \( u = \frac{3}{2} \), \( u(3-u) = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \). Hence, \( f = \frac{9}{4} \times \frac{1}{3} = \frac{9}{12} = 0.75 \).

Step 6: Choose the correct answer

According to the calculations, the maximum possible focal length is \( 0.75 \mathrm{~m} \). Hence, the answer is option (c).

Key Concepts

Focal Length Calculation

When dealing with the focal length of a convex lens, it's important to understand how the focal length relates to the placement of objects and images. In this exercise, the lens is used to project an image of a bulb across a room, 3 meters apart from the object to the image. Calculating focal length accurately involves a combination of distances in optics, mainly the object distance (\( u \)) and the image distance (\( v \)).

To calculate the maximum focal length, you can use the arrangement where the object distance added to the image distance equals the distance between the walls: \( u + v = 3 \text{ meters} \). This relation is crucial for deriving a suitable mathematical expression. Understanding this relationship allows us to find the maximum focal length, ensuring the image is precisely formed on the opposite wall.

Lens Formula

The lens formula is fundamental to understanding image formation through a lens. It is expressed as\( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), where \( f \) is the focal length, \( u \) is the object distance, and \( v \) is the image distance. Given that these values can vary, the formula offers a way to compute unknown distances if other values are known.

For problems like projecting a bulb's image onto a wall, modifying the equation helps fit practical scenarios. By considering \( u + v = 3 \) meters (the total distance between walls), you can express \( v \) as \( v = 3 - u \). Substituting this back into the lens formula aligns the equation to context-specific problems. This type of simple manipulation of the equation helps you quickly model systems and find the necessary focal length value.

Image Formation

Understanding how images are formed through a convex lens demands a basic grasp of what's happening to light rays. When you place an object in front of a lens, the light rays travel through the lens and converge to form an image. The nature and type of the image—real or virtual, magnified or diminished—are dictated by the object's distance in relation to the lens's focal length.

In our exercise here, the light from the bulb on one side of the room needs to converge on the opposite wall. This is a case of a real image formation since the image is projected onto an actual surface. Calculating the precise distances, using the lens equation helps configure the lens to get this desired real image just at the wall across, ensuring clarity and focus.

Optics Concepts

Convex lenses are a fundamental topic in optics, a branch of physics that deals with light's behavior and properties. A lens's ability to focus or spread out light beams is a result of its shape and refractive index. As with our convex lens, these lenses are thicker in the center than at the edges, which allows them to converge light rays to a focal point.

Understanding some essential optics principles, such as refraction and focal points, aids in comprehending how lenses work. In refraction, light bends as it enters a different medium, such as from air into the glass of a lens, changing direction based on the lens's curvature. In designing optical setups or solving related problems like the bulb and wall scenario, knowing these basics ensures better command over how focal length and image distances hinge on these fundamental concepts.