An object is placed a distance of 200 cm from a combination of two equal lenses of equal power of -1 D- The distance of the image of the object system is-33-33 cm-33-33 cm66-67 cm40- cm

Question

Toppr Admin · Accepted Answer

P-P1-P2-2Df-12m-50cm1u-1v-1f1v-1f-x2212-1u-150-x2212-1-x2212-200-140v-40cm

An object is placed at a distance of $200 c m$ from a combination of two equal lenses of equal power of $+ 1 D$ . The distance of the image of the object for system is:
$33.33 c m$
$- 33.33 c m$
$66.67 c m$
$40 c m$

$P = P_{1} + P_{2} = 2 D$
$f = \frac{1}{2} m = 50 c m$
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
$= \frac{1}{50} - \frac{1}{- 200} = \frac{1}{40}$
$v = 40 c m$

An object is placed at a distance of 200 cm from a combination of two equal lenses of equal power of +1 D. The distance of the image of the object for system is:33.33 cm−33.33 cm66.67 cm40cm

P=P1+P2=2Df=12m=50cm1u+1v=1f1v=1f−1u=150−1−200=140v=40cm

An object is placed at a distance of $200 c m$ from a combination of two equal lenses of equal power of $+ 1 D$ . The distance of the image of the object for system is:
$33.33 c m$
$- 33.33 c m$
$66.67 c m$
$40 c m$

$P = P_{1} + P_{2} = 2 D$
$f = \frac{1}{2} m = 50 c m$
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$
$= \frac{1}{50} - \frac{1}{- 200} = \frac{1}{40}$
$v = 40 c m$