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: Begin by determining the lens's focal length using the lens formula:
   \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where:
   • \( f \) represents the focal length
   • \( v \) is the image distance
   • \( u \) is the object distance
2. Step 2: Before adding liquid (lens in air), we have:
   • \( u = -36 \, \text{cm} \)
   • \( v = 45 \, \text{cm} \)
   Calculate the focal length in air:
   \[ \frac{1}{f_{\text{air}}} = \frac{1}{45} - \frac{1}{-36} \] Solve for \( f_{\text{air}} \) to find the focal length in air.
3. Step 3: Introducing the liquid changes the lens's effective focal length due to its refractive index. The new image distance is:
   • \( v' = 48 \, \text{cm} \)
   Apply the lens formula again, using the new object distance \( u' \) and the focal length in liquid \( f_{\text{liquid}} \):
   \[ n = \frac{f_{\text{liquid}}}{f_{\text{air}}} \] Substituting the values, the refractive index is:
   \[ n = 1.366 \]