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

Steps to Solve

1. Identify the given values The object length ($h_o$) is (5 , \text{cm}), the object distance ($u$) is (-20 , \text{cm}) (it is negative because it is on the opposite side of the lens), and the focal length ($f$) is (30 , \text{cm}).

2. Use the lens formula The lens formula is given by

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

Substituting the values we have:

$$ \frac{1}{30} = \frac{1}{v} - \frac{1}{(-20)} $$

3. Rearranging the equation Rearranging the equation to find (v):

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

4. Finding a common denominator The common denominator for (30) and (20) is (60). Rewrite the fractions:

$$ \frac{1}{v} = \frac{2}{60} + \frac{3}{60} = \frac{5}{60} $$

5. Calculate (v) Taking the reciprocal gives:

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

6. Determine the magnification The magnification ((m)) of the lens is given by

$$ m = \frac{h_i}{h_o} = -\frac{v}{u} $$

Substituting the values:

$$ m = -\frac{12}{-20} = \frac{12}{20} = \frac{3}{5} $$

7. Find the image length Now, using the magnification to find the image height ((h_i)):

$$ h_i = m \cdot h_o = \frac{3}{5} \times 5 , \text{cm} = 3 , \text{cm} $$