An object of height 3 cm is placed at a distance 30 cm in front of a convex lens of focal length 20 cm. The height of the image will be:

Correct option is A

To solve this problem, we can use the lens formula and the magnification formula.

Given:

• Object height,$$h_o = 3 \, \text{cm}$$​​
• Object distance, u=−30cm(Negetive because the object is placed in front of the lens)
• Focal length of the convex lens, f=20cm (positive for a convex lens)

The lens formula is:

​$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$​​

Where:

• f is the focal length,
• v is the image distance,
• u is the object distance.

Substituting the known values:

$$\frac{1}{20} = \frac{1}{v} - \frac{1}{-30}$$​​

​$$\frac{1}{20} = \frac{1}{v} + \frac{1}{30}$$​​

Now, solve for $$\frac{1}{v}:$$​​

​$$\frac{1}{v} = \frac{1}{20} - \frac{1}{30}$$​​

Find a common denominator:

​$$\frac{1}{v} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60}$$​​

Thus:

     v=60cm

The magnification M is given by the formula:

$$M = \frac{h_i}{h_o} = \frac{v}{u}$$​

Substitute the values for v and u:

         $$M = \frac{60}{-30} = -2$$​

The negative sign indicates that the image is inverted. Now, calculate the image height $$h_i:$$​​

​$$M = \frac{h_i}{h_o} \quad \Rightarrow \quad -2 = \frac{h_i}{3}$$​​

Solving for $$h_i:$$​​

​$$h_i = -2 \times 3 = -6 \, \text{cm}$$​​

So, the image height is 6 cm (inverted).

The height of the image is 6 cm, and the correct option is:

(a) 6 cm.