A concave mirror has the form of a hemisphere with a radius of `R=60cm` A thin layer of an unknown transparent liquid is poured into the mirror ,The mirror-liquids system forms one real Image and another real image is formed by mirror alone ,with the sources in a certain position .One of them coincides with the source and the other is at distance of `l=30 cm` from source.find the possible value(s) refractive index `mu`of the liquid.

Explanation

To find the possible value(s) of the refractive index \( \mu \) of the liquid in a concave mirror system, we will follow these steps:

### Step 1: Understand the System
The concave mirror has a radius of curvature \( R = 60 \, \text{cm} \). The focal length \( f \) of the mirror is given by:
\[ f = -\frac{R}{2} = -\frac{60}{2} = -30 \, \text{cm} \]

### Step 2: Analyze the Image Formation
When the mirror is filled with a liquid, the system forms two real images:
1. One image coincides with the object.
2. The other image is at a distance \( l = 30 \, \text{cm} \) from the object.

Let’s denote the object distance for the mirror alone as \( u = -60 \, \text{cm} \) (since it is in front of the mirror) and the image distance as \( v = -30 \, \text{cm} \) when the liquid is present.

### Step 3: Apply the Mirror Formula
Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values for the mirror alone:
\[ \frac{1}{-30} = \frac{1}{v} + \frac{1}{-60} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-30} + \frac{1}{60} \]
Calculating:
\[ \frac{1}{v} = -\frac{2}{60} + \frac{1}{60} = -\frac{1}{60} \]
Thus, \( v = -60 \, \text{cm} \).

### Step 4: Analyze the Liquid's Effect
When the liquid is present, the effective focal length of the system changes. The effective power \( P_{\text{effective}} \) of the system can be expressed as:
\[ P_{\text{effective}} = P_L + P_M \]
Where \( P_M \) is the power of the mirror and \( P_L \) is the power of the liquid lens.

### Step 5: Calculate the Power of the Mirror
The power of the mirror is given by:
\[ P_M = \frac{1}{f_M} = -\frac{1}{30} \]

### Step 6: Calculate the Power of the Liquid
Using the distances for the liquid-filled system, we can express the effective power:
\[ \frac{1}{F_{\text{effective}}} = \frac{1}{v} + \frac{1}{u} \]
Substituting \( v = -30 \, \text{cm} \) and \( u = -60 \, \text{cm} \):
\[ \frac{1}{F_{\text{effective}}} = \frac{1}{-30} + \frac{1}{-60} \]
Calculating gives:
\[ \frac{1}{F_{\text{effective}}} = -\frac{2}{60} - \frac{1}{60} = -\frac{3}{60} = -\frac{1}{20} \]
Thus, \( F_{\text{effective}} = -20 \, \text{cm} \).

### Step 7: Relate Power of Liquid to Refractive Index
Using the lens maker's formula:
\[ \frac{1}{F_L} = \mu - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Assuming \( R_1 \) is infinite (flat surface) and \( R_2 = -60 \, \text{cm} \):
\[ \frac{1}{F_L} = \mu - 1 \left( 0 + \frac{1}{60} \right) \]
Thus:
\[ \frac{1}{-20} = \mu - 1 \left( -\frac{1}{60} \right) \]
Rearranging gives:
\[ \mu - 1 = -\frac{1}{20} \cdot -60 = \frac{3}{10} \]
So:
\[ \mu = \frac{3}{10} + 1 = \frac{13}{10} \]

### Step 8: Conclusion
The possible value of the refractive index \( \mu \) of the liquid is:
\[ \mu = 1.3 \]