(a) Two lenses A and B have power of (i) +2 D and (ii) –4 D respectively. What is the nature and focal length of each lens?
(b) An object is placed at a distance of 100 cm from each of the above lenses A and Calculate (i) image distance, and (ii) magnification, in each of the two cases.
(a) PA = +2D
\(f_A\) = \(\frac{1}{P_A}\) = \(\frac{1}{2}\)
= +0.5m = +50 cm
Lens A is a convex lens.
PB = -4D
\(f_B\) = \(\frac{1}{P_B}\) = \(\frac{1}{-4}\)
= -0.25 m = -25 cm
Lens B is a concave lens.
(b) Case 1: For lens A
fA = +50 cm
uA = -100 cm
\(\frac{1}{f_A}\) = \(\frac{1}{v_A}\) - \(\frac{1}{u_A}\)
\(\frac{1}{v_A}\) = \(\frac{1}{f_A}\) + \(\frac{1}{u_A}\)
= \(\frac{1}{50}\) + \(\frac{1}{-100}\)
= \(\frac{1}{100}\)
Image distance, vA = 100 cm
Magnification, mA = \(\frac{v_A}{u_A}\)
= \(\frac{100}{-100}\) = -1
Case 1: For lens B
fB = -25 cm
uB = -100 cm
\(\frac{1}{f_B}\) = \(\frac{1}{v_B}\) - \(\frac{1}{u_B}\)
\(\frac{1}{v_B}\) = \(\frac{1}{f_B}\) + \(\frac{1}{u_B}\)
= \(\frac{1}{-25}\) + \(\frac{1}{-100}\)
= \(\frac{-5}{100}\)
Image distance, vB = -20 cm
Magnification, mB = \(\frac{v_B}{u_B}\)
= \(\frac{-20}{-100}\) = +0.2