A biconvex lens has a radius of curvature of magnitude `20cm`. Which one of the following options describes best the image formed of an object of height `2cm` place `30 cm` from the lens ?

Options

A. Virtual, upright, height= `0.5cm`
B. Real, inverted, height=`4 cm`
C. Real, inverted, height =`1cm`
D. Virtual, upright, height=`1 cm`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given parameters
- Radius of curvature (R) = 20 cm
- Object distance (u) = -30 cm (the negative sign indicates that the object is placed on the same side as the incoming light)
- Height of the object (h) = 2 cm

### Step 2: Calculate the focal length (f) of the lens using the lensmaker's formula
The lensmaker's formula for a biconvex lens is given by:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a biconvex lens:
- \( n = \frac{3}{2} \) (refractive index of glass)
- \( R_1 = +20 \, \text{cm} \) (first surface, convex)
- \( R_2 = -20 \, \text{cm} \) (second surface, concave)

Substituting the values:
\[ \frac{1}{f} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{20} - \left(-\frac{1}{20}\right) \right) \]
\[ = \frac{1}{2} \left( \frac{1}{20} + \frac{1}{20} \right) = \frac{1}{2} \left( \frac{2}{20} \right) = \frac{1}{20} \]
Thus, the focal length \( f = 20 \, \text{cm} \).

### Step 3: Use the lens formula to find the image distance (v)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{20} = \frac{1}{v} - \frac{1}{-30} \]
This simplifies to:
\[ \frac{1}{20} = \frac{1}{v} + \frac{1}{30} \]
To solve for \( \frac{1}{v} \):
\[ \frac{1}{v} = \frac{1}{20} - \frac{1}{30} \]
Finding a common denominator (60):
\[ \frac{1}{v} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60} \]
Thus, \( v = 60 \, \text{cm} \).

### Step 4: Calculate the magnification (m)
The magnification formula is given by:
\[ m = \frac{h'}{h} = \frac{v}{u} \]
Where \( h' \) is the height of the image and \( h \) is the height of the object. Substituting the values:
\[ m = \frac{60}{-30} = -2 \]
This means the height of the image \( h' \) can be calculated as:
\[ h' = m \cdot h = -2 \cdot 2 = -4 \, \text{cm} \]

### Step 5: Interpret the results
The negative sign of the height indicates that the image is inverted. Since the image distance \( v \) is positive, it indicates that the image is real.

### Final Conclusion
The image formed is real, inverted, and has a height of 4 cm.