Question

The apparent depth of a needle lying in a water beaker is found to be 9 cm. If water is replaced by a liquid of refractive index 1.5, then the apparent depth of needle will be (μ of water is 4/3)

• A. 10 cm
• B. 9 cm
• C. 12 cm
• D. 8 cm
• E. 7 cm

Answer 1

The Correct Option is D

We are given: 

• Apparent depth in water = 9 cm
• Refractive index of water, \( \mu_{\text{water}} = \frac{4}{3} \)
• Refractive index of new liquid, \( \mu_{\text{liquid}} = 1.5 \)

Step 1: Find the real depth using the water's refractive index: \[ \text{Real depth} = \mu_{\text{water}} \times \text{Apparent depth} = \frac{4}{3} \times 9 = 12 \, \text{cm} \]

Step 2: Use the real depth to find the new apparent depth: \[ \text{Apparent depth in liquid} = \frac{\text{Real depth}}{\mu_{\text{liquid}}} = \frac{12}{1.5} = 8 \, \text{cm} \]

Final Answer: \( \boxed{8 \, \text{cm} } \)

Answer 2

Given:

• Apparent depth in water, \( d_{\text{app}}^{\text{water}} = 9 \, \text{cm} \) 
• Refractive index of water, \( \mu_{\text{water}} = \frac{4}{3} \)
• Refractive index of the liquid, \( \mu_{\text{liquid}} = 1.5 \)

We know that the apparent depth is related to the refractive index and actual depth:

\[ d_{\text{app}} \propto \mu \]

\p>Thus, we have the relationship between the apparent depth in water and the apparent depth in the liquid:

 

\[ \frac{d_{\text{app}}^{\text{water}}}{d_{\text{app}}^{\text{liquid}}} = \frac{\mu_{\text{liquid}}}{\mu_{\text{water}}} \]

Substituting the given values:

\[ \frac{9}{d_{\text{app}}^{\text{liquid}}} = \frac{1.5}{\frac{4}{3}} \]

Simplifying the equation:

\[ \frac{9}{d_{\text{app}}^{\text{liquid}}} = \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125 \]

Solving for \( d_{\text{app}}^{\text{liquid}} \):

\[ d_{\text{app}}^{\text{liquid}} = \frac{9}{1.125} = 8 \, \text{cm} \]

Therefore, the apparent depth in the liquid is \( 8 \, \text{cm} \).