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Question

A point source in the air is placed at a distance of 40 cm in front of a spherical convex glass surface (μ 2 = 1.5) of radius of curvature 10 cm. The image of the source is formed at a distance of _______ from the surface in the direction of incident  light.

The correct answer is

60 cm

Understanding how light behaves when it passes through different media is fundamental in optics. This problem involves a point source and a spherical convex glass surface, requiring the application of the spherical refraction formula to determine the image location.

Spherical Surface Refraction Principles

When light travels from one medium to another through a spherical surface, its path changes. This phenomenon is known as refraction. The relationship between object distance, image distance, radii of curvature, and refractive indices is described by the spherical refraction formula. This formula is crucial for solving problems involving image formation by spherical surfaces, whether convex or concave.

  • Point Source: The object emitting light is considered a single point.
  • Spherical Convex Glass Surface: A curved surface that bulges outwards, made of glass with a specific refractive index.
  • Refractive Index (\(\mu\)): A measure of how much the speed of light is reduced in a medium. Air has a refractive index of approximately 1, while glass has a higher refractive index, typically around 1.5.
  • Radius of Curvature (R): The radius of the sphere of which the surface is a part. For a convex surface, the radius of curvature is taken as positive when the light is incident from the left and the center of curvature is to the right, following the Cartesian sign convention.

Optical Parameters for Image Formation

Let's identify the given values from the question for accurate calculation of the image position formed by the spherical convex glass surface:

  • Refractive index of the first medium (air), \(\mu_1 = 1\).
  • Refractive index of the second medium (glass), \(\mu_2 = 1.5\).
  • Object distance (distance of the point source from the surface), \(u = -40\) cm. The negative sign indicates that the object is placed in front of the surface, as per the sign convention where light travels from left to right.
  • Radius of curvature of the convex glass surface, \(R = +10\) cm. For a convex surface, the center of curvature is on the side of the refracting medium, so R is positive according to the Cartesian sign convention.

Image Distance Calculation for Spherical Surface

To find the image distance (\(v\)), we use the formula for refraction at a single spherical surface:

\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]

Now, substitute the known values into the formula:

\[ \frac{1.5}{v} - \frac{1}{-40} = \frac{1.5 - 1}{10} \]

Simplify the equation:

\[ \frac{1.5}{v} + \frac{1}{40} = \frac{0.5}{10} \]

\[ \frac{1.5}{v} + \frac{1}{40} = \frac{1}{20} \]

Isolate the term with \(v\):

\[ \frac{1.5}{v} = \frac{1}{20} - \frac{1}{40} \]

Find a common denominator for the right side:

\[ \frac{1.5}{v} = \frac{2}{40} - \frac{1}{40} \]

\[ \frac{1.5}{v} = \frac{1}{40} \]

Solve for \(v\):

\[ v = 1.5 \times 40 \]

\[ v = 60 \text{ cm} \]

Convex Surface Image Location Analysis

The calculated image distance \(v = +60\) cm is positive. In the Cartesian sign convention, a positive image distance means that the image is formed on the side where the refracted light exits the surface. For a spherical convex glass surface, if the object is in air and light passes into glass, a positive image distance implies the image is formed inside the glass, on the side of the incident light's propagation direction. This indicates that the image formed is a real image.

Spherical Refraction Summary

The image of the point source placed in front of the spherical convex glass surface is formed at a distance of 60 cm from the surface in the direction of the incident light. This result aligns with the principles of refraction at spherical surfaces and typical image formation scenarios for convex refracting surfaces.

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Important Questions from Optics

  1. Which one of the following colours may be obtained by combining green and red colours?

  2. Which of the following are the primary colours of light?

  3. Directions: The following items consist of two statements, Statement I and Statement II. You are to examine these two statements carefully and select the answers to these items using the code given below:

    Statement I:  Diamond is very bright.

    Statement II: Diamond has very low refractive index

  4. A non-SI unit called 'nit' is the unit of which of the following photometric quantities used to measure a multitude of light intensity?

  5. Which among the following is used as a reflector in search lights?

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