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A double convex lens has focal length \(25 \mathrm{~cm}\). The radius of curvature of one of the surfaces is double of the other. Find the radii, if the refractive index of the material of the lens is \(1 \cdot 5\).

Short Answer

Expert verified
Radii of curvature are 6.25 cm and 12.5 cm.

Step by step solution

01

Understanding the lens maker's formula

The lens maker's formula relates the focal length of a lens to the refractive index and the radii of curvature of the two surfaces of the lens. It is given by: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]where \(f\) is the focal length, \(n\) is the refractive index, and \(R_1\) and \(R_2\) are the radii of curvature of the lens surfaces.
02

Substituting known values

We have a focal length \( f = 25 \) cm, and refractive index \( n = 1.5 \). Let's denote the radius of curvature of the first surface as \( R \) and the second surface as \( 2R \), since one radius is twice the other. Substitute the values into the formula:\[ \frac{1}{25} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{2R} \right) \].
03

Simplifying the equation

Simplify the term \( \left( \frac{1}{R} - \frac{1}{2R} \right) \):\[ = \frac{2 - 1}{2R} = \frac{1}{2R} \]Substitute back into the equation:\[ \frac{1}{25} = 0.5 \cdot \frac{1}{2R} \]which simplifies to:\[ \frac{1}{25} = \frac{1}{4R} \].
04

Solving for the radius

To find \( R \), set the equation \( \frac{1}{25} = \frac{1}{4R} \):\[ 4R = 25 \]Solving for \( R \), we get:\[ R = \frac{25}{4} = 6.25 \text{ cm} \]
05

Determining both radii of curvature

Since one radius is double the other, if \( R = 6.25 \) cm, then the other radius is:\[ 2R = 2 \times 6.25 = 12.5 \text{ cm} \].Thus, the radii of curvature are \( 6.25 \text{ cm} \) and \( 12.5 \text{ cm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Focal Length Calculation
The focal length of a lens is a measure of how strongly it converges or diverges light. When we have two opposite curved surfaces, like those of a lens, the focal length determines the point where light rays meet after passing through the lens. Calculating this focal length involves understanding how the lens' material bends light and the geometry of its surfaces.

The lens maker's formula is crucial for this calculation. It gives us a way to find the focal length using the refractive index of the material and the radii of curvature of its surfaces. The formula is:
  • \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
Here, \(f\) is the focal length. By knowing the refractive index \(n\) and the curved surfaces' radii \(R_1\) and \(R_2\), we can calculate \(f\). This relationship helps us design lenses with desired focal properties, enabling sharp images in glasses, cameras, and microscopes.
Refractive Index
The refractive index, \(n\), is a dimensionless number that indicates how much a medium, like the material of a lens, can bend light. This bending ability is critical in controlling the focal length of the lens. A higher refractive index means that light slows down more and bends significantly when passing into the material.

For example, a refractive index of \(1.5\) tells us that light travels 1.5 times slower in the lens material than in a vacuum. This reduces the speed of light significantly, altering its path and thereby influencing how a lens focuses light.

In practical terms, choosing materials with different refractive indices allows manufacturers to create lenses of various strengths and uses. From eyeglasses to complex camera lenses, understanding and utilizing this property is essential for optical engineering.
Radii of Curvature
The radii of curvature \(R_1\) and \(R_2\) are the radii of the spheres from which the lens surfaces are part. These values have a direct impact on how much a lens can converge or diverge light. A larger radius implies a flatter surface, which bends light less, whereas a smaller radius indicates a steeper curve, bending light more.

In a double convex lens, the radii can be different for each surface. As described in the exercise, one radius may be double the other. This difference helps in achieving specific focal properties.

By using the lens maker's formula, you can see how these radii link to the focal length and refractive index:
  • \( \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \)
Understanding the relationship between these factors is key to designing lenses with precise focusing capabilities, essential in numerous technological applications.

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Most popular questions from this chapter

A pin of length \(2 \cdot 00 \mathrm{~cm}\) is placed perpendicular to the principal axis of a converging lens. An inverted image of size \(1 \cdot 00 \mathrm{~cm}\) is formed at a distance of \(40 \cdot 0 \mathrm{~cm}\) from the pin. Find the focal length of the lens and its distance from the pin.

A converging lens of focal length \(12 \mathrm{~cm}\) and a diverging mirror of focal length \(7 \cdot 5 \mathrm{~cm}\) are placed \(5 \cdot 0 \mathrm{~cm}\) apart with their principal axes coinciding. Where should an object be placed so that its image falls on itself?

A gun of mass \(M\) fires a bullet of mass \(m\) with a horizontal speed \(V\). The gun is fitted with a concave mirror of focal length \(f\) facing towards the receding bullet. Find the speed of separation of the bullet and the image just after the gun was fired.

A point source is placed at a depth \(h\) below the surface of water (refractive index \(=\mu\) ). (a) Show that light escapes through a circular area on the water surface with its centre directly above the point source. (b) Find the angle subtended by a radius of the area on the source.

A small block of mass \(m\) and a concave mirror of radius \(R\) fitted with a stand lie on a smooth horizontal table with a separation \(d\) between them. The mirror together with its stand has a mass \(m .\) The block is pushed at \(t=0\) towards the mirror so that it starts moving towards the mirror at a constant speed \(V\) and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time \(td / V\).

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