Two equi-convex lenses of focal length `30cm` and `70cm,` made of material of refractive index`=1.5,` are held in contact coaxially by a rubber band round their edges.A liquidof refractive index `1.3` is introduced in the space between the lenses filling it completely. Find the position of the image of a luminous point object placed on teh axis of the combination lens at a distance of `90cm` from it.

Explanation

To solve the problem step-by-step, we need to follow these steps:

### Step 1: Identify the given data
- Focal lengths of the lenses:
- \( f_1 = 30 \, \text{cm} \)
- \( f_2 = 70 \, \text{cm} \)
- Refractive index of the lenses: \( n = 1.5 \)
- Refractive index of the liquid: \( n_{liquid} = 1.3 \)
- Distance of the object from the lens combination: \( u = -90 \, \text{cm} \) (negative as per the sign convention)

### Step 2: Calculate the focal length of the liquid-filled lens
Using the lens maker's formula for the combination of the two lenses with the liquid in between, we can find the effective focal length \( f_3 \) of the liquid-filled lens.

The lens maker's formula is given by:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For the first lens (focal length \( f_1 \)):
- \( R_1 = +30 \, \text{cm} \) (convex)
- \( R_2 = -30 \, \text{cm} \) (for the second surface of the first lens)

For the second lens (focal length \( f_2 \)):
- \( R_3 = +70 \, \text{cm} \) (convex)
- \( R_4 = -70 \, \text{cm} \) (for the second surface of the second lens)

Using the formula for the liquid-filled lens:
\[ f_3 = \frac{1}{(n_{liquid} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \]
Substituting the values:
\[ f_3 = \frac{1}{(1.3 - 1) \left( \frac{1}{-30} - \frac{1}{70} \right)} \]
Calculating:
\[ f_3 = \frac{1}{0.3 \left( -\frac{1}{30} - \frac{1}{70} \right)} \]
Finding a common denominator:
\[ -\frac{1}{30} - \frac{1}{70} = -\frac{7 + 3}{210} = -\frac{10}{210} = -\frac{1}{21} \]
Thus,
\[ f_3 = \frac{1}{0.3 \left( -\frac{1}{21} \right)} = \frac{-21}{0.3} = -70 \, \text{cm} \]

### Step 3: Calculate the equivalent focal length of the combination
Using the formula for the equivalent focal length of the combination of lenses:
\[ \frac{1}{f_{equiv}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \]
Substituting the values:
\[ \frac{1}{f_{equiv}} = \frac{1}{30} + \frac{1}{70} + \frac{1}{-70} \]
This simplifies to:
\[ \frac{1}{f_{equiv}} = \frac{1}{30} \]
Thus,
\[ f_{equiv} = 30 \, \text{cm} \]

### Step 4: Use the lens formula to find the image position
Using the lens formula:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Substituting \( u = -90 \, \text{cm} \) and \( f = 30 \, \text{cm} \):
\[ \frac{1}{v} - \left(-\frac{1}{90}\right) = \frac{1}{30} \]
This simplifies to:
\[ \frac{1}{v} + \frac{1}{90} = \frac{1}{30} \]
Finding a common denominator:
\[ \frac{1}{v} = \frac{1}{30} - \frac{1}{90} = \frac{3 - 1}{90} = \frac{2}{90} = \frac{1}{45} \]
Thus,
\[ v = 45 \, \text{cm} \]

### Conclusion
The position of the image is \( 45 \, \text{cm} \) to the right of the lens combination.

---