A thin lens of refractive index `1.5` has focal length of `15 cm` in air. When the lens is placed is a medium of refractive index (4)/(3), its focal length will become …..cm.

Explanation

To find the new focal length of a thin lens when it is placed in a medium of a different refractive index, we can use the lens maker's formula. Here are the step-by-step calculations:

### Step 1: Write down the lens maker's formula
The lens maker's formula is given by:

\[ \frac{1}{f} = \left(\frac{\mu_L}{\mu_{surrounding}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Where:
- \( f \) is the focal length of the lens,
- \( \mu_L \) is the refractive index of the lens,
- \( \mu_{surrounding} \) is the refractive index of the surrounding medium,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Determine the values for the first scenario (in air)
Given:
- \( \mu_L = 1.5 \)
- \( f = 15 \, \text{cm} \)

Using the formula, we can rearrange it to find \( \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \):

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

Calculating \( \frac{1.5}{1} - 1 \):

\[ \frac{1.5}{1} - 1 = 0.5 \]

So we have:

\[ \frac{1}{15} = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

### Step 3: Solve for \( \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Rearranging gives:

\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{15 \times 0.5} = \frac{1}{7.5} \]

### Step 4: Determine the values for the second scenario (in a medium of refractive index \( \frac{4}{3} \))
Now, when the lens is placed in a medium with \( \mu_{surrounding} = \frac{4}{3} \) or approximately \( 1.33 \):

Using the lens maker's formula again:

\[ \frac{1}{f'} = \left(\frac{\mu_L}{\mu_{surrounding}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Substituting the known values:

\[ \frac{1}{f'} = \left(\frac{1.5}{\frac{4}{3}} - 1\right) \left(\frac{1}{7.5}\right) \]

Calculating \( \frac{1.5}{\frac{4}{3}} \):

\[ \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125 \]

So:

\[ \frac{1}{f'} = (1.125 - 1) \left(\frac{1}{7.5}\right) = 0.125 \left(\frac{1}{7.5}\right) \]

### Step 5: Calculate \( f' \)
Now we can calculate \( f' \):

\[ \frac{1}{f'} = \frac{0.125}{7.5} = \frac{0.125}{7.5} = \frac{1}{60} \]

Thus:

\[ f' = 60 \, \text{cm} \]

### Final Answer:
The focal length of the lens when placed in a medium of refractive index \( \frac{4}{3} \) will be **60 cm**.

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