A double convex lens of focal length 30 cm is made of glass. When it is immersed in a liquid of refractive index 1.4, the focal length is found to be 126 cm. The critical angle between glass and the liquid is

Options

A. `sin^(-1)(3/4)`
B. `sin^(-1)(4/5)`
C. `sin^(-1)(7/13)`
D. `sin^(-1)(7/8)`

Answer

To find the critical angle between the glass and the liquid, we can follow these steps:

### Step 1: Identify the refractive indices
We are given:
- The refractive index of glass (μg) = 1.6
- The refractive index of the liquid (μl) = 1.4

### Step 2: Calculate the relative refractive index
The relative refractive index (μ) between the glass and the liquid can be calculated using the formula:
\[ \mu = \frac{\mu_g}{\mu_l} \]
Substituting the values:
\[ \mu = \frac{1.6}{1.4} \]

### Step 3: Simplify the relative refractive index
Calculating the above expression:
\[ \mu = \frac{1.6}{1.4} = \frac{16}{14} = \frac{8}{7} \]

### Step 4: Use the critical angle formula
The critical angle (θc) can be found using the formula:
\[ \sin \theta_c = \frac{1}{\mu} \]
Substituting the value of μ:
\[ \sin \theta_c = \frac{1}{\frac{8}{7}} = \frac{7}{8} \]

### Step 5: Calculate the critical angle
To find the critical angle, we take the inverse sine:
\[ \theta_c = \sin^{-1}\left(\frac{7}{8}\right) \]

### Conclusion
Thus, the critical angle between the glass and the liquid is:
\[ \theta_c = \sin^{-1}\left(\frac{7}{8}\right) \]