A glass sphere of refractive index `1.5` forms the real image of object `O` at point `I` as shown in the figure when kept in air of refractive index `1`. If the value of `x` is `kR`. Find `k`

Explanation

To solve the problem, we need to analyze the behavior of light as it passes through the glass sphere and forms images at specific points. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have a glass sphere with a refractive index of \( n_2 = 1.5 \) in air with a refractive index of \( n_1 = 1 \). The object \( O \) forms a real image \( I \) at a distance \( x \) from the center of the sphere.

### Step 2: First Refraction
When light travels from the object \( O \) to the first surface of the sphere, we can denote the radius of the sphere as \( R \). The object distance \( u \) for the first refraction (from the image point \( I \)) is taken as \( u = -R \) (the negative sign indicates that the object is on the same side as the incoming light).

Using the lens maker's formula for refraction at a spherical surface:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
Substituting the values:
\[ \frac{1.5}{v} - \frac{1}{-R} = \frac{1.5 - 1}{R} \]
This simplifies to:
\[ \frac{1.5}{v} + \frac{1}{R} = \frac{0.5}{R} \]
Rearranging gives:
\[ \frac{1.5}{v} = \frac{0.5}{R} - \frac{1}{R} = -\frac{0.5}{R} \]
Thus,
\[ v = -3R \]

### Step 3: Second Refraction
Now, the image \( I_1 \) acts as the object for the second refraction at the second surface of the sphere. The distance from the center of the sphere to the image \( I_1 \) is \( 3R \), but we need to consider the total distance from the second surface, which is \( 5R \) (since it is \( 3R + 2R \)).

Now, we have:
\[ u = -5R \]
Using the lens maker's formula again for the second refraction:
\[ \frac{n_3}{x} - \frac{n_2}{u} = \frac{n_3 - n_2}{R} \]
Substituting the values:
\[ \frac{1}{x} - \frac{1.5}{-5R} = \frac{1 - 1.5}{R} \]
This simplifies to:
\[ \frac{1}{x} + \frac{1.5}{5R} = -\frac{0.5}{R} \]
Rearranging gives:
\[ \frac{1}{x} = -\frac{0.5}{R} - \frac{1.5}{5R} \]
Finding a common denominator:
\[ \frac{1}{x} = -\frac{2.5}{5R} - \frac{1.5}{5R} = -\frac{4}{5R} \]
Thus,
\[ x = -\frac{5R}{4} \]

### Step 4: Finding \( k \)
Given that \( x = kR \), we can equate:
\[ kR = -\frac{5R}{4} \]
Thus,
\[ k = -\frac{5}{4} \]

### Final Answer
The value of \( k \) is \( -\frac{5}{4} \).

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