Problem 56 - Master Resource Book in JEE Main Physics

Question

Double-convex lenses are to be manufactured from a glass of refractive index \(1.55\), with both faces of the same radius of curvature. What is the radius of curvature required, if the foeal length is to be \(20 \mathrm{~cm} ?\) [NCERT] (a) \(18 \mathrm{~m}\) (b) \(22 \mathrm{~cm}\) (c) \(17 \mathrm{~cm}\) (d) \(26 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The radius of curvature required is 22 cm.

Step 1: Understand Lens Maker's Formula

The Lens Maker's Formula is used to relate the focal length of a lens to the radii of curvature of its two surfaces and its refractive index. For a lens in air, the formula is: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]where \(f\) is the focal length of the lens, \(n\) is the refractive index, and \(R_1\) and \(R_2\) are the radii of curvature of the two surfaces. For a double-convex lens with equal radii, \(R_1 = R\) and \(R_2 = -R\).

Step 2: Substitute Known Values

Given that the refractive index \(n = 1.55\) and focal length \(f = 20 \text{ cm}\), and both radii of curvature are equal, we substitute these into the Lens Maker's Formula:\[\frac{1}{20} = (1.55 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right)\]This simplifies to:\[\frac{1}{20} = 0.55 \left( \frac{2}{R} \right)\]

Step 3: Solve for Radius of Curvature

Rearrange the simplified equation to isolate \(R\):\[\frac{1}{20} = \frac{1.1}{R}\]This implies:\[R = 1.1 \times 20\]Calculate the product:\[R = 22 \text{ cm}\]

Step 4: Verify and Conclude

Check that the calculations match the options provided. The calculated radius of curvature is 22 cm, which matches option (b). Therefore, both logic and calculations support the correctness of this result.

Key Concepts

Refractive Index

The refractive index is a measure of how much a material can bend light. This property is key in the study of optics and crucial for understanding how lenses work. A refractive index ( ) is defined as the ratio of the speed of light in a vacuum to the speed of light in a given material. When a material has a high refractive index, it bends light more.

• If the refractive index is greater than 1, the medium is optically denser than a vacuum or air, which is precisely 1.00.
• In our exercise, the refractive index of the glass used in making the lens is 1.55, which means light travels slower within the glass than it does in air.

This strongly affects the lens's ability to focus light, making it a central aspect of the lens maker’s formula. Understanding how refractive index influences light bending helps in predicting how lenses can broaden or narrow the field of view and focus of an image.

Radius of Curvature

The radius of curvature describes the degree of curvature of a lens's surface. Imagine slicing a sphere through the center; the radius of curvature is the radius of that sphere. For a double-convex lens, we have two surfaces, each potentially having a different radius. However, in our problem scenario, both radii are equal due to the lens being symmetrical.

• The symbol for radius of curvature is typically represented as R.
• In the context of the lens maker's formula, when both sides of the lens have equal radii, it simplifies the equation significantly. Here, the specific assumption is that one side is positive and the other is negative, mathematically represented as R and -R.

This concept is crucial because it directly influences the calculations for determining the focal length, and in turn, the practical usage of lenses for various optical applications, such as in cameras and glasses.

Focal Length

The focal length is a fundamental characteristic of lenses that indicates the distance from the lens to the focal point, where parallel rays of light meet after passing through the lens. This distance determines how powerfully a lens converges or diverges light. In simple terms, it affects how "zoomed in" or "wide angle" the lens is.

• A shorter focal length means a stronger converging power, leading to a magnified image.
• A longer focal length results in a wider field of view, common in landscape photography.

In the lens maker’s formula, focal length (f) is inversely proportional to the radii of curvature and the refractive index of the lens material. In our Ncvert problem, we know f is 20 cm, allowing us to calculate the radius of curvature for the special case where both radii are equal. This deep understanding of focal length empowers users to select the right lens type for various tasks, from reading glasses to telescope lenses.