Question

A thin convex lens of refractive index 1.5 has a focal length of 10 cm in air. When the lens is immersed in a fluid,its focal length becomes 70 cm. The refractive index of the fluid is

• A. 1.33
• B. 1.6
• C. 1.25
• D. 1.45
• E. 1.4

Answer 1

Answer: (E)

\[ \frac{1}{f} = (n_{\text{lens}} - n_{\text{medium}}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

where:
- \( f \) is the focal length of the lens,
- \( n_{\text{lens}} \) is the refractive index of the lens material,
- \( n_{\text{medium}} \) is the refractive index of the surrounding medium,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

Given:
- \( n_{\text{lens}} = 1.5 \),
- \( f_{\text{air}} = 10 \text{ cm} \),
- \( f_{\text{fluid}} = 70 \text{ cm} \).

First, we find the curvature term using the focal length in air (\( n_{\text{medium}} = 1 \)):

\[ \frac{1}{10} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

\[ \frac{1}{10} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10 \times 0.5} = \frac{1}{5} \]

Next, we use the same curvature term with the focal length in the fluid to find the refractive index of the fluid \( n_{\text{fluid}} \):

\[ \frac{1}{70} = (1.5 - n_{\text{fluid}}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substitute \( \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{5} \):

\[ \frac{1}{70} = (1.5 - n_{\text{fluid}}) \times \frac{1}{5} \]

\[ \frac{1}{70} = \frac{1.5 - n_{\text{fluid}}}{5} \]

\[ 5 \times \frac{1}{70} = 1.5 - n_{\text{fluid}} \]

\[ \frac{1}{14} = 1.5 - n_{\text{fluid}} \]

\[ n_{\text{fluid}} = 1.5 - \frac{1}{14} \]

\[ n_{\text{fluid}} = 1.5 - 0.0714 \]

\[ n_{\text{fluid}} = 1.4286 \]

Thus, the refractive index of the fluid is approximately \( 1.42 \).

So The correct answer is Option(E):1.4 nearest matching answer

Answer 2

Step 1: Recall the lens maker's formula.

The focal length of a thin convex lens is given by the lens maker's formula:

\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right), \]

where:

• \( f \) is the focal length of the lens,
• \( n \) is the refractive index of the lens material relative to the medium (air or fluid),
• \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens.

 

In air, the refractive index of the lens relative to air is \( n_{\text{air}} = \frac{n_{\text{lens}}}{n_{\text{air}}} = 1.5 \), where \( n_{\text{lens}} = 1.5 \) and \( n_{\text{air}} = 1 \).

In the fluid, the refractive index of the lens relative to the fluid is \( n_{\text{fluid}} = \frac{n_{\text{lens}}}{n_{\text{fluid}}} \), where \( n_{\text{fluid}} \) is the refractive index of the fluid.

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Step 2: Write the lens maker's formula for both cases.

(i) In air:

\[ \frac{1}{f_{\text{air}}} = (n_{\text{air}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right). \]

Substitute \( f_{\text{air}} = 10 \, \text{cm} \) and \( n_{\text{air}} = 1.5 \):

\[ \frac{1}{10} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right). \]

Simplify:

\[ \frac{1}{10} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right). \]

\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{5}. \]

(ii) In the fluid:

\[ \frac{1}{f_{\text{fluid}}} = (n_{\text{fluid}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right). \]

Substitute \( f_{\text{fluid}} = 70 \, \text{cm} \) and \( \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{5} \):

\[ \frac{1}{70} = (n_{\text{fluid}} - 1) \cdot \frac{1}{5}. \]

Simplify:

\[ n_{\text{fluid}} - 1 = \frac{1}{70} \cdot 5 = \frac{5}{70} = \frac{1}{14}. \]

\[ n_{\text{fluid}} = 1 + \frac{1}{14} = \frac{15}{14} \approx 1.07. \] ---

Final Answer: The refractive index of the fluid is approximately \( \mathbf{1.4} \), which corresponds to option \( \mathbf{(E)} \).