An object of length 5 cm is placed perpendicular to the principal axis (f = 30 cm) at 20 cm from its optic centre. Find the image length and its properties.

Steps to Solve

1. Identify Given Information

   We have the following values:

   • Object length ($h_o = 5 , \text{cm}$)
   • Distance from the optical center ($u = -20 , \text{cm}$; negative because it's measured against the direction of the incident light)
   • Focal length of the lens ($f = 30 , \text{cm}$)

2. Apply the Lens Formula

   The lens formula is given by: $$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$

   Substituting the known values: $$ \frac{1}{30} = \frac{1}{v} - \frac{1}{-20} $$

3. Rearranging the Lens Formula

   Rearranging the equation to isolate $v$: $$ \frac{1}{v} = \frac{1}{30} + \frac{1}{20} $$

4. Finding Common Denominator and Simplifying

   The common denominator for 30 and 20 is 60: $$ \frac{1}{v} = \frac{2}{60} + \frac{3}{60} = \frac{5}{60} $$

   Thus: $$ v = \frac{60}{5} = 12 , \text{cm} $$

5. Calculate the Magnification

   The magnification ($m$) is given by: $$ m = -\frac{v}{u} = -\frac{12}{-20} = \frac{12}{20} = \frac{3}{5} $$

6. Find Image Length

   The image length ($h_i$) can be calculated using: $$ h_i = m \times h_o = \frac{3}{5} \times 5 , \text{cm} = 3 , \text{cm} $$