An object is 8cm high. It is desired to form a real image 4cm high at 20cm from the lens. The focal length and power of lends are

Options

A. convex mirror with focal length f=40cm
B. convex mirror with focal length f=20cm
C. convex mirror with focal length f=-40cm
D. convex mirror with focal length f=-240cm

Answer

To solve the problem step by step, we will follow the principles of optics, specifically using the lens formula and magnification concepts.

### Step 1: Identify the given parameters
- Height of the object (h_o) = 8 cm
- Height of the image (h_i) = 4 cm (real image)
- Distance of the image from the lens (V) = 20 cm

### Step 2: Calculate the magnification (m)
The magnification (m) can be calculated using the formula:
\[ m = \frac{h_i}{h_o} \]
Substituting the known values:
\[ m = \frac{4 \, \text{cm}}{8 \, \text{cm}} = \frac{1}{2} \]

### Step 3: Relate magnification to object distance (U)
The magnification can also be expressed in terms of object distance (U) and image distance (V):
\[ m = \frac{V}{U} \]
Since we have already calculated \( m \) as \( \frac{1}{2} \):
\[ \frac{1}{2} = \frac{20 \, \text{cm}}{U} \]
Cross-multiplying gives:
\[ U = 2 \times 20 \, \text{cm} = 40 \, \text{cm} \]

### Step 4: Use the lens formula to find the focal length (F)
The lens formula is given by:
\[ \frac{1}{F} = \frac{1}{V} - \frac{1}{U} \]
Substituting the values of V and U:
\[ \frac{1}{F} = \frac{1}{20 \, \text{cm}} - \frac{1}{40 \, \text{cm}} \]

### Step 5: Calculate the right-hand side
Finding a common denominator (which is 40):
\[ \frac{1}{F} = \frac{2}{40} - \frac{1}{40} = \frac{1}{40} \]

### Step 6: Solve for Focal Length (F)
Taking the reciprocal gives:
\[ F = 40 \, \text{cm} \]

### Step 7: Calculate the power of the lens (P)
The power (P) of a lens is given by:
\[ P = \frac{1}{F \, (\text{in meters})} \]
Converting F from cm to meters:
\[ F = 40 \, \text{cm} = 0.4 \, \text{m} \]
Thus,
\[ P = \frac{1}{0.4} = 2.5 \, \text{D} \]

### Final Results
- Focal Length (F) = 40 cm
- Power (P) = 2.5 D