A thin convex lens from a real Image of a certain object 'p' times it size .The size of real image become 'q' times that of object when the lens is moved nearer to the object by a distance 'a' find focal length of the lens?

Options

A. ` (|a|)/(| p - q | ) `
B. ` ( apq ) / ( | p - q | ) `
C. Both 1 & 2
D. None

Explanation

To solve the problem step by step, we will use the lens formula and magnification concepts. Let's break it down:

### Step 1: Understanding the Initial Setup
Let the object distance be \( u = -x \) (negative as per the sign convention) and the image distance be \( v \). The magnification \( m \) is given as \( p \), which means:
\[ m = \frac{v}{u} = p \]
From this, we can express \( v \) in terms of \( u \):
\[ v = p \cdot u = p \cdot (-x) = -px \]

### Step 2: Applying the Lens Formula
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( v \) and \( u \):
\[ \frac{1}{f} = \frac{1}{-px} - \frac{1}{-x} \]
This simplifies to:
\[ \frac{1}{f} = -\frac{1}{px} + \frac{1}{x} = \frac{1}{x} - \frac{1}{px} = \frac{p - 1}{px} \]
Let’s label this as Equation (1):
\[ \frac{1}{f} = \frac{p - 1}{px} \tag{1} \]

### Step 3: Analyzing the New Position of the Lens
When the lens is moved nearer to the object by a distance \( a \), the new object distance becomes:
\[ u' = -x - a \]
The new image distance \( v' \) can be expressed in terms of the new magnification \( q \):
\[ v' = -q \cdot (x + a) \]

### Step 4: Applying the Lens Formula Again
Using the lens formula again for the new distances:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
Substituting the new values:
\[ \frac{1}{f} = \frac{1}{-q(x + a)} - \frac{1}{- (x + a)} \]
This simplifies to:
\[ \frac{1}{f} = -\frac{1}{q(x + a)} + \frac{1}{x + a} = \frac{1}{x + a} - \frac{1}{q(x + a)} = \frac{q - 1}{q(x + a)} \]
Let’s label this as Equation (2):
\[ \frac{1}{f} = \frac{q - 1}{q(x + a)} \tag{2} \]

### Step 5: Equating the Two Expressions for \( \frac{1}{f} \)
From Equations (1) and (2), we equate:
\[ \frac{p - 1}{px} = \frac{q - 1}{q(x + a)} \]

### Step 6: Cross-Multiplying to Solve for \( x \)
Cross-multiplying gives:
\[ (q - 1)px = (p - 1)q(x + a) \]
Expanding and rearranging:
\[ (q - 1)px = (p - 1)qx + (p - 1)qa \]
\[ (qp - q - p + 1)x = (p - 1)qa \]
Thus,
\[ x = \frac{(p - 1)qa}{(qp - q - p + 1)} \]

### Step 7: Substituting Back to Find \( f \)
Now substitute \( x \) back into Equation (1) to find \( f \):
\[ \frac{1}{f} = \frac{p - 1}{p \left(\frac{(p - 1)qa}{(qp - q - p + 1)}\right)} \]
Simplifying gives:
\[ f = \frac{pqa}{qp - q - p + 1} \]

### Final Result
Thus, the focal length \( f \) of the lens is:
\[ f = \frac{pq a}{q - p} \]