Question

Light from a point source in air falls on a spherical glass surface (refractive index, \( \mu = 1.5 \) and radius of curvature \( R = 50 \) cm). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is 1cm.

Hint:

Apply the formula for refraction at a spherical surface, \( \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \). Be careful with the sign conventions for \( u \), \( v \), and \( R \). For a convex surface, \( R \) is positive when light goes from a rarer to a denser medium. Real objects have negative \( u \), and real images formed on the opposite side of the refracting surface have positive \( v \). Ensure consistent units throughout the calculation and convert the final answer to the required unit.

Answer 1

Correct Answer: 4

We will use the formula for refraction at a spherical surface: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \] where: \( \mu_1 \) is the refractive index of the first medium (air) = 1 
\( \mu_2 \) is the refractive index of the second medium (glass) = 1.5 
\( u \) is the object distance from the spherical surface (to be found) 
\( v \) is the image distance from the spherical surface = 200 cm (positive as the image is formed in the second medium) 
\( R \) is the radius of curvature of the spherical surface = +50 cm (positive as the surface is convex to the incident light) 
Substituting the given values into the formula: \[ \frac{1.5}{200} - \frac{1}{u} = \frac{1.5 - 1}{50} \] \[ \frac{1.5}{200} - \frac{1}{u} = \frac{0.5}{50} \] \[ \frac{1.5}{200} - \frac{1}{u} = \frac{1}{100} \] \[ \frac{3}{400} - \frac{1}{u} = \frac{1}{100} \] \[ \frac{1}{u} = \frac{3}{400} - \frac{1}{100} = \frac{3}{400} - \frac{4}{400} = -\frac{1}{400} \] \[ u = -400 \, \text{cm} \] The negative sign indicates that the object is real and located on the side from which the light is incident. 
The magnitude of the distance of the light source from the glass surface is \( |u| = 400 \, \text{cm} \). 
To convert this distance to meters, we divide by 100: \[ |u| = \frac{400}{100} \, \text{m} = 4 \, \text{m} \]

Answer 2

The problem asks for the distance of a point light source from a spherical glass surface, given the refractive index of the glass, its radius of curvature, and the position where the image is formed inside the glass.

Concept Used:

The relationship between the object distance (\(u\)), image distance (\(v\)), refractive indices of the two media (\(n_1\) and \(n_2\)), and the radius of curvature (\(R\)) for refraction at a single spherical surface is given by the formula:

\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

We will use the Cartesian sign convention, where the pole of the spherical surface is the origin. Distances measured in the direction of the incident light are taken as positive, and those measured against it are negative.

Step-by-Step Solution:

Step 1: List the given values according to the sign convention. Light travels from air (medium 1) to glass (medium 2).

• Refractive index of air, \(n_1 = 1\)
• Refractive index of glass, \(n_2 = \mu = 1.5\)
• Radius of curvature, \(R = +50\) cm (The surface is convex towards the object, so R is positive).
• Image distance, \(v = +200\) cm (The image is formed inside the glass, which is in the direction of light, so v is positive).
• Object distance, \(u = ?\)

Step 2: Substitute the known values into the refraction formula.

\[ \frac{1.5}{+200} - \frac{1}{u} = \frac{1.5 - 1}{+50} \]

Step 3: Simplify the right-hand side of the equation.

\[ \frac{1.5 - 1}{50} = \frac{0.5}{50} = \frac{1}{100} \]

Step 4: The equation now becomes:

\[ \frac{1.5}{200} - \frac{1}{u} = \frac{1}{100} \]

Step 5: Isolate the term containing \(u\).

\[ \frac{1}{u} = \frac{1.5}{200} - \frac{1}{100} \]

Step 6: Find a common denominator to solve for \(\frac{1}{u}\).

\[ \frac{1}{u} = \frac{1.5}{200} - \frac{2}{200} \] \[ \frac{1}{u} = \frac{1.5 - 2}{200} = \frac{-0.5}{200} \]

Step 7: Simplify the fraction and solve for \(u\).

\[ \frac{1}{u} = -\frac{0.5}{200} = -\frac{1}{400} \] \[ u = -400 \text{ cm} \]

The negative sign confirms that the object is located in front of the refracting surface, in the first medium (air). The magnitude of the distance is 400 cm.

Step 8: Convert the magnitude of the distance from centimeters to meters as required by the question.

\[ \text{Distance} = 400 \text{ cm} = \frac{400}{100} \text{ m} = 4 \text{ m} \]

The magnitude of the distance of the light source from the glass surface is 4 m.