Problem 59 - Master Resource Book in JEE Main Physics

Question

The radius of the convex surface of plane-convex lens is \(20 \mathrm{~cm}\) and the refractive index of the material of the lens is 1.5. The focal length of the lens is (a) \(30 \mathrm{~cm}\) (b) \(50 \mathrm{~cm}\) (c) \(20 \mathrm{~cm}\) (d) \(40 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The correct answer is (d) 40 cm.

Step 1: Understand Lens Formula

For a lens, the formula relating the focal length \( f \), the refractive index \( n \), and the radius of curvature \( R \) of the lens is given by the lens maker's equation: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]. For a plane-convex lens, one side is flat, so \( R_2 = \infty \), simplifying to \[ \frac{1}{f} = (n - 1) \frac{1}{R} \]. We need to calculate the focal length \( f \).

Step 2: Plug in Known Values

In the simplified lens maker's formula \( \frac{1}{f} = (n-1) \frac{1}{R} \), plug in \( n = 1.5 \) and \( R = 20 \text{ cm} \). This gives us: \( \frac{1}{f} = (1.5 - 1) \frac{1}{20} \).

Step 3: Solve for Focal Length

Calculate the change in refractive index by subtracting 1 from \( n \): \( 1.5 - 1 = 0.5 \). Therefore, the equation becomes \( \frac{1}{f} = 0.5 \times \frac{1}{20} \). Simplify this to get \( \frac{1}{f} = \frac{0.5}{20} = \frac{1}{40} \).

Step 4: Calculate Final Answer

Now solve for \( f \) by taking the reciprocal of both sides: \( f = 40 \text{ cm} \).

Key Concepts

Lens Maker's Equation

The lens maker's equation is fundamental in optics. It establishes a relationship between the focal length of a lens, its refractive index, and the curvature of its surfaces. For any lens, this equation is expressed as: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Where:

• \( f \) is the focal length, the distance where parallel rays converge after passing through the lens.
• \( n \) stands for the refractive index, a measure of how much light is bent or refracted when entering the lens.
• \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens.

For a plane-convex lens, like the one in our problem, one side is flat, making its radius of curvature \( R_2 \) infinite. Consequently, the equation simplifies to \[ \frac{1}{f} = (n - 1) \frac{1}{R} \]. This simplification makes it easier to calculate lenses' focal lengths where one surface is planar.

Focal Length

The focal length of a lens is a critical concept in optics. It is defined as the distance from the lens at which parallel rays of light converge. This fundamental distance determines how a lens focuses light and, subsequently, images. The focal length is influenced by:

• Lens curvature: Highly curved lenses have shorter focal lengths and converge light more rapidly.
• Refractive index: Lenses made from materials with a higher refractive index can also shorten the focal length.
• Lens maker's equation: As seen, the equation directly links these variables to determine \( f \).

In practical usage, a shorter focal length means a wider field of view, while longer focal lengths narrow the field, focusing more on distant objects.

Refractive Index

The refractive index is a number that describes how fast light travels through a material. It is an essential parameter in optical systems, greatly influencing lens behavior. Formally defined, it is the ratio of the speed of light in a vacuum to its speed in the material. This is given by: \[ n = \frac{c}{v} \] where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light within the material.Several factors influence the refractive index:

• Material composition: Denser materials generally have higher refractive indices.
• Wavelength of light: The refractive index can vary with the wavelength due to dispersion, meaning it is different for various colors.

The refractive index is vital for determining how light is bent and focused by lenses, affecting their focal length based on the lens maker’s equation.