Question

When a convex lens is placed above an empty tank, the image of a mark at the bottom of the tank, which is 45 cm from the lens is formed 36 cm above the lens. When a liquid is poured in the tank to a depth of 40 cm, the distance of the image of the mark above the lens is 48 cm. The refractive index of the liquid is:

• A. 1.353
• B. 1.544
• C. 1.472
• D. 1.366

Hint:

To calculate the refractive index of a medium, use the lens formula to determine the focal length in both air and the medium, then take their ratio.

Answer 1

The Correct Option is D

1. Step 1: First, we need to calculate the focal length of the lens using the lens formula. The lens formula is:
   \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where:
   • \( f \) is the focal length
   • \( v \) is the image distance
   • \( u \) is the object distance
2. Step 2: When the lens is in air (before any liquid is added), we have:
   • \( u = -36 \, \text{cm} \)
   • \( v = 45 \, \text{cm} \)
   The focal length in air can be calculated as:
   \[ \frac{1}{f_{\text{air}}} = \frac{1}{45} - \frac{1}{-36} \] Solving for \( f_{\text{air}} \), we get the focal length in air.
3. Step 3: When the liquid is poured into the tank, the refractive index of the liquid alters the effective focal length of the lens. The new image distance is given as:
   • \( v' = 48 \, \text{cm} \)
   Using the same lens formula, but with the new object distance \( u' \) and focal length in the liquid \( f_{\text{liquid}} \), we calculate:
   \[ n = \frac{f_{\text{liquid}}}{f_{\text{air}}} \] Substituting the values, the refractive index is found to be:
   \[ n = 1.366 \]