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 need to find the focal length and power of the lens given the height of the object, the height of the image, and the distance of the image from the lens.

### Step 1: Write down the given data
- Height of the object (h_o) = 8 cm
- Height of the image (h_i) = 4 cm
- Distance of the image from the lens (v) = 20 cm

### Step 2: Calculate the magnification (m)
The magnification (m) is given by the formula:
\[ m = \frac{h_i}{h_o} \]

Substituting the 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, we can set up the equation:
\[ \frac{1}{2} = -\frac{20}{u} \]

### Step 4: Solve for object distance (u)
Rearranging the equation gives:
\[ u = -\frac{20 \times 2}{1} = -40 \, \text{cm} \]

### Step 5: Apply the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

Substituting the values of v and u:
\[ \frac{1}{f} = \frac{1}{20} - \frac{1}{-40} \]

### Step 6: Simplify the equation
Calculating the right-hand side:
\[ \frac{1}{f} = \frac{1}{20} + \frac{1}{40} \]

Finding a common denominator (which is 40):
\[ \frac{1}{f} = \frac{2}{40} + \frac{1}{40} = \frac{3}{40} \]

### Step 7: Calculate the focal length (f)
Taking the reciprocal:
\[ f = \frac{40}{3} \, \text{cm} \approx 13.33 \, \text{cm} \]

### Step 8: Calculate the power (P) of the lens
The power of the lens is given by:
\[ P = \frac{1}{f \, (\text{in meters})} \]

Converting focal length to meters:
\[ f = \frac{40}{3} \, \text{cm} = \frac{40}{300} \, \text{m} = \frac{2}{15} \, \text{m} \]

Now calculating the power:
\[ P = \frac{1}{\frac{2}{15}} = \frac{15}{2} = 7.5 \, \text{D} \]

### Final Answer
- Focal length (f) = 13.33 cm
- Power (P) = 7.5 D