An optical system consists of two convergent lenses with focal length `f_1=20 cm and f_2=10 cm.` The distance between the lenses is `d=30 cm.` An object is placed at a distance of `30 cm` from the first lens. At what distance from the second lens will the images be obtained ?

Explanation

To solve the problem step by step, we will analyze the optical system consisting of two convergent lenses.

### Step 1: Identify the parameters
We have two convergent lenses:
- Focal length of the first lens, \( f_1 = 20 \, \text{cm} \)
- Focal length of the second lens, \( f_2 = 10 \, \text{cm} \)
- Distance between the lenses, \( d = 30 \, \text{cm} \)
- Object distance from the first lens, \( u_1 = -30 \, \text{cm} \) (negative because the object is on the same side as the incoming light)

### Step 2: Use the lens formula for the first lens
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
For the first lens (L1):
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \]
Substituting the values:
\[ \frac{1}{20} = \frac{1}{v_1} - \frac{1}{(-30)} \]
This simplifies to:
\[ \frac{1}{20} = \frac{1}{v_1} + \frac{1}{30} \]

### Step 3: Solve for \( v_1 \)
To solve for \( v_1 \), first find a common denominator (LCM of 20 and 30 is 60):
\[ \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \]
Thus, we have:
\[ \frac{3}{60} = \frac{1}{v_1} + \frac{2}{60} \]
Rearranging gives:
\[ \frac{1}{v_1} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60} \]
Therefore:
\[ v_1 = 60 \, \text{cm} \]

### Step 4: Determine the object distance for the second lens
The image formed by the first lens (L1) acts as the object for the second lens (L2). The distance from the first lens to the second lens is \( d = 30 \, \text{cm} \). Therefore, the object distance for the second lens is:
\[ u_2 = d - v_1 = 30 - 60 = -30 \, \text{cm} \]
(Note: It is negative because it's on the same side as the incoming light for the second lens.)

### Step 5: Use the lens formula for the second lens
Now we apply the lens formula for the second lens (L2):
\[ \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \]
Substituting the values:
\[ \frac{1}{10} = \frac{1}{v_2} - \frac{1}{(-30)} \]
This simplifies to:
\[ \frac{1}{10} = \frac{1}{v_2} + \frac{1}{30} \]

### Step 6: Solve for \( v_2 \)
Finding a common denominator (LCM of 10 and 30 is 30):
\[ \frac{1}{10} = \frac{3}{30}, \quad \frac{1}{30} = \frac{1}{30} \]
Thus, we have:
\[ \frac{3}{30} = \frac{1}{v_2} + \frac{1}{30} \]
Rearranging gives:
\[ \frac{1}{v_2} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} \]
Therefore:
\[ v_2 = 15 \, \text{cm} \]

### Conclusion
The final image is formed at a distance of \( 15 \, \text{cm} \) from the second lens.