A double convex lens has faces of radii of curvature 30 cm each. The refractive index of the material of the lens is 1.5. What is the focal length of this lens when immersed is carbondisulphide of refractive index 1.6 ?

Options

A. `-120 cm `
B. `-180 cm `
C. `-200 cm `
D. `-240 cm `

Explanation

To find the focal length of a double convex lens immersed in carbon disulfide, we can follow these steps:

### Step 1: Determine the Refractive Index of the Lens with Respect to Carbon Disulfide
The refractive index of the lens material (glass) with respect to carbon disulfide can be calculated using the formula:
\[ n_{g/c} = \frac{n_g}{n_c} \]
where:
- \( n_g \) = refractive index of the lens material = 1.5
- \( n_c \) = refractive index of carbon disulfide = 1.6

Calculating this gives:
\[ n_{g/c} = \frac{1.5}{1.6} = \frac{15}{16} \]

### Step 2: Use the Lensmaker's Formula
The lensmaker's formula for a lens in a medium other than air is given by:
\[ \frac{1}{f} = (n_{g/c} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a double convex lens:
- \( R_1 = +30 \, \text{cm} \) (positive for the first surface)
- \( R_2 = -30 \, \text{cm} \) (negative for the second surface)

Substituting these values into the formula:
\[ \frac{1}{f} = \left( \frac{15}{16} - 1 \right) \left( \frac{1}{30} - \left(-\frac{1}{30}\right) \right) \]

### Step 3: Simplify the Equation
Calculating \( n_{g/c} - 1 \):
\[ n_{g/c} - 1 = \frac{15}{16} - 1 = \frac{15 - 16}{16} = -\frac{1}{16} \]
Now substituting into the formula:
\[ \frac{1}{f} = -\frac{1}{16} \left( \frac{1}{30} + \frac{1}{30} \right) = -\frac{1}{16} \left( \frac{2}{30} \right) = -\frac{1}{16} \cdot \frac{1}{15} = -\frac{1}{240} \]

### Step 4: Find the Focal Length
Taking the reciprocal to find \( f \):
\[ f = -240 \, \text{cm} \]

### Conclusion
The focal length of the lens when immersed in carbon disulfide is \( -240 \, \text{cm} \).

---