A point object is placed at 30 cm from a convex glass lens `(mu_(g) = (3)/(2))` of focal length 20 cm. The final image of object will be formed at infinity if

Options

A. another concave lens of focal length 60 cm is placed in contact with the previous lens
B. another convex lens of focal length 60 cm is placed at a distance of 30 cm from the first lens
C. the whole system is immersed in a liquid of refractive index 4/3
D. the whole system is immersed in a liquid of refractive index 9/8

Answer

To solve the problem step by step, we need to analyze the situation involving a convex lens and the conditions under which the final image of the object will be formed at infinity.

### Step 1: Understand the Given Data
- Object distance (u) = -30 cm (the negative sign indicates that the object is on the same side as the incoming light).
- Focal length of the convex lens (f) = 20 cm.
- The refractive index of the glass lens (µg) = 3/2.

### Step 2: Determine the Condition for the Image to be at Infinity
For the image to be formed at infinity, the object must be placed at the focus of the lens. Therefore, we need to find the equivalent focal length (F) of the lens system such that:
\[ F = u \]
This means:
\[ F = -30 \text{ cm} \]

### Step 3: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
Where \( f_1 \) is the focal length of the convex lens (20 cm) and \( f_2 \) is the focal length of the second lens (unknown).

### Step 4: Substitute Known Values
Substituting the known values into the lens formula:
\[ \frac{1}{-30} = \frac{1}{20} + \frac{1}{f_2} \]

### Step 5: Solve for \( f_2 \)
Rearranging the equation to solve for \( \frac{1}{f_2} \):
\[ \frac{1}{f_2} = \frac{1}{-30} - \frac{1}{20} \]
Finding a common denominator (60):
\[ \frac{1}{f_2} = \frac{-2}{60} - \frac{3}{60} = \frac{-5}{60} \]
Thus:
\[ f_2 = -12 \text{ cm} \]

### Step 6: Identify the Type of Lens
Since \( f_2 \) is negative, this indicates that the second lens is a concave lens with a focal length of 12 cm.

### Step 7: Check Options for Immersion in Liquid
Now, we need to check the conditions under which the system is immersed in a liquid. The lens maker's formula is:
\[ \frac{1}{F} = \frac{µg}{µm - 1} \cdot \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where \( µm \) is the refractive index of the medium.

### Step 8: Calculate the Effective Focal Length in Different Mediums
Using the lens maker's formula, we can derive the conditions for the system to have an effective focal length of -30 cm when immersed in a liquid.

### Step 9: Solve for \( µm \)
By setting up the equations for the refractive index of the medium, we can find that:
\[ µm = \frac{9}{8} \]

### Conclusion
The final answers are:
- A concave lens of focal length 60 cm is placed in contact with the convex lens.
- The system is immersed in a liquid with a refractive index of \( \frac{9}{8} \).