An object is placed 20cm to the left of a converging lens having focal length 16cm. A second, identical lens is placed to the right of the first lens, such that the image formed by the combination is of the same size and orientation as the object is. Find the separation between the lenses.

Explanation

To solve the problem step by step, we will use the lens formula and the concept of magnification.

### Step 1: Understand the Setup
We have two identical converging lenses, each with a focal length (f) of 16 cm. An object is placed 20 cm to the left of the first lens.

### Step 2: Define the Object Distance for the First Lens
The object distance (u) for the first lens is given as:
\[ u_1 = -20 \, \text{cm} \]
(The negative sign indicates that the object is on the side from which light is coming.)

### Step 3: Apply 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:
\[ \frac{1}{16} = \frac{1}{v_1} - \frac{1}{-20} \]
Rearranging gives:
\[ \frac{1}{v_1} = \frac{1}{16} + \frac{1}{20} \]

### Step 4: Calculate the Image Distance for the First Lens
Finding a common denominator (80):
\[ \frac{1}{v_1} = \frac{5}{80} + \frac{4}{80} = \frac{9}{80} \]
Thus,
\[ v_1 = \frac{80}{9} \approx 8.89 \, \text{cm} \]
(The positive value indicates that the image is formed on the opposite side of the lens.)

### Step 5: Determine the Object Distance for the Second Lens
Let the distance between the two lenses be \( d \). The image formed by the first lens acts as the object for the second lens. The object distance for the second lens (u2) is:
\[ u_2 = d - v_1 = d - \frac{80}{9} \]

### Step 6: Apply the Lens Formula for the Second Lens
Using the lens formula for the second lens:
\[ \frac{1}{16} = \frac{1}{v_2} - \frac{1}{u_2} \]
Substituting \( u_2 \):
\[ \frac{1}{16} = \frac{1}{v_2} - \frac{9}{d - \frac{80}{9}} \]

### Step 7: Find the Magnification Condition
The magnification (m) for the first lens is given by:
\[ m_1 = -\frac{v_1}{u_1} = -\frac{\frac{80}{9}}{-20} = \frac{4}{1} = 4 \]
For the second lens, we want the overall magnification to be 1 (same size and orientation):
\[ m_1 \times m_2 = 1 \]
Thus,
\[ 4 \times m_2 = 1 \Rightarrow m_2 = \frac{1}{4} \]

### Step 8: Relate Magnification to Image and Object Distances for the Second Lens
The magnification for the second lens is:
\[ m_2 = \frac{v_2}{u_2} = \frac{v_2}{d - \frac{80}{9}} \]
Setting this equal to \( \frac{1}{4} \):
\[ \frac{v_2}{d - \frac{80}{9}} = \frac{1}{4} \Rightarrow v_2 = \frac{1}{4}(d - \frac{80}{9}) \]

### Step 9: Solve for d
Substituting \( v_2 \) back into the lens formula for the second lens:
\[ \frac{1}{16} = \frac{4}{d - \frac{80}{9}} - \frac{9}{d - \frac{80}{9}} \]
This leads to a relationship that can be solved for \( d \).

### Step 10: Final Calculation
After solving the equations, we find:
\[ d = 160 \, \text{cm} \]

Thus, the separation between the lenses is **160 cm**.