When a biconvex lens of glass having refractive index `1.47` is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index.

Options

A. less than that of glass
B. equal to that of glass
C. less than one
D. greater than that of glass

Explanation

To determine the refractive index of the liquid in which a biconvex lens of glass (with a refractive index of 1.47) is dipped, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We know that when the biconvex lens is dipped in a liquid, it behaves like a plane sheet of glass. This implies that the lens's focal length becomes infinite.

2. **Use the Lensmaker's Formula**: The lensmaker's formula relates the focal length (f) of a lens to its refractive indices and radii of curvature:
\[ \frac{1}{f} = \left(\mu_g - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
where:
- \( \mu_g \) = refractive index of the glass (1.47)
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

3. **Set Focal Length to Infinity**: Since the lens acts like a plane sheet of glass, we set the focal length \( f \) to infinity:
\[ \frac{1}{\infty} = 0 \]

4. **Rearranging the Formula**: This leads to:
\[ 0 = \left(\mu_g - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]
For this equation to hold true, either \( \mu_g - 1 = 0 \) or \( \frac{1}{R_1} - \frac{1}{R_2} = 0 \).

5. **Analyzing the Conditions**: Since the lens is biconvex, the radii of curvature \( R_1 \) and \( R_2 \) are not equal (they have opposite signs). Therefore, \( \frac{1}{R_1} - \frac{1}{R_2} \neq 0 \).

6. **Setting the Refractive Indices Equal**: Thus, we conclude that:
\[ \mu_g - 1 = 0 \implies \mu_g = \mu_L \]
where \( \mu_L \) is the refractive index of the liquid.

7. **Final Calculation**: Since \( \mu_g = 1.47 \), we find:
\[ \mu_L = 1.47 \]

### Conclusion:
The refractive index of the liquid must be \( 1.47 \).