Three lenses in contact have a combined focal length of 12 cm. When the third lens is removed, the combined focal length is `(60)/(7)` cm. The third lens is

Options

A. A converging lens of focal length 30 cm
B. A converging lens of foca length 60 cm
C. A diverging lens of focal length 30 cm
D. A diverging lens of focal length 60 cm

Answer

To solve the problem, we need to find the focal length of the third lens when three lenses are in contact and their combined focal length is given. Let's break this down step by step.

### Step 1: Understand the combined focal length of three lenses
When three lenses (f1, f2, f3) are in contact, the formula for the combined focal length (F) is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \]
From the problem, we know that:
\[ F = 12 \text{ cm} \]
Thus, we can write:
\[ \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} = \frac{1}{12} \]

### Step 2: Understand the combined focal length when the third lens is removed
When the third lens (f3) is removed, the combined focal length of the remaining two lenses (f1 and f2) is given as:
\[ F' = \frac{60}{7} \text{ cm} \]
So we can write:
\[ \frac{1}{f_1} + \frac{1}{f_2} = \frac{7}{60} \]

### Step 3: Set up the equations
Now we have two equations:
1. \(\frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} = \frac{1}{12}\)
2. \(\frac{1}{f_1} + \frac{1}{f_2} = \frac{7}{60}\)

### Step 4: Substitute the second equation into the first
From the second equation, we can substitute \(\frac{1}{f_1} + \frac{1}{f_2}\) into the first equation:
\[ \frac{7}{60} + \frac{1}{f_3} = \frac{1}{12} \]

### Step 5: Solve for \(\frac{1}{f_3}\)
Rearranging the equation gives:
\[ \frac{1}{f_3} = \frac{1}{12} - \frac{7}{60} \]

### Step 6: Find a common denominator and calculate
To subtract the fractions, we need a common denominator. The least common multiple of 12 and 60 is 60. Thus:
\[ \frac{1}{12} = \frac{5}{60} \]
Now substituting:
\[ \frac{1}{f_3} = \frac{5}{60} - \frac{7}{60} = \frac{-2}{60} \]
This simplifies to:
\[ \frac{1}{f_3} = -\frac{1}{30} \]

### Step 7: Find \(f_3\)
Taking the reciprocal gives:
\[ f_3 = -30 \text{ cm} \]

### Conclusion
The focal length of the third lens is \(-30\) cm, indicating that it is a concave lens (or diverging lens).