Two thin lenses have a combined power of +9 D. When they are separated by a distance of 20 cm, their equivalent power becomes +(27/5) D. Their individual powers (in diopters) are

Question

Two thin lenses have a combined power of +9 D. When they are separated by a distance of 20 cm, their equivalent power becomes +(27/5) D. Their individual powers (in diopters) are (A) 1,8 (B) 2,7 (C) 3,6 (D) 4,5

Tardigrade Admin · Accepted Answer

Given, P1+P2=9 P=P1+P2-d P1 P2 (27/5)=9-(20/100) × P1 P2 Hence, the above equation is correct for P1=3 and P3=6

Q. Two thin lenses have a combined power of $+ 9 D$ . When they are separated by a distance of $20 c m$ , their equivalent power becomes $+ \frac{27}{5} D$ . Their individual powers (in diopters) are

1,8

2,7

3,6

4,5

Given,
$P_{1} + P_{2} = 9$
$P = P_{1} + P_{2} - d P_{1} P_{2}$
$\frac{27}{5} = 9 - \frac{20}{100} \times P_{1} P_{2}$
Hence, the above equation is correct for
$P_{1} = 3$ and $P_{3} = 6$

Q. Two thin lenses have a combined power of +9D. When they are separated by a distance of 20cm, their equivalent power becomes +527​D. Their individual powers (in diopters) are

1,8

2,7

3,6

4,5

Given, P1​+P2​=9 P=P1​+P2​−dP1​P2​ 527​=9−10020​×P1​P2​ Hence, the above equation is correct for P1​=3 and P3​=6

Q. Two thin lenses have a combined power of $+ 9 D$ . When they are separated by a distance of $20 c m$ , their equivalent power becomes $+ \frac{27}{5} D$ . Their individual powers (in diopters) are

Given,
$P_{1} + P_{2} = 9$
$P = P_{1} + P_{2} - d P_{1} P_{2}$
$\frac{27}{5} = 9 - \frac{20}{100} \times P_{1} P_{2}$
Hence, the above equation is correct for
$P_{1} = 3$ and $P_{3} = 6$