Two lenses have 10 D power each and they are separated by a distance. Beyond which distance does the power of combination changes from positive to negative?

Question

Two lenses have 10 D power each and they are separated by a distance. Beyond which distance does the power of combination changes from positive to negative? (A) 5 cm (B) 10 cm (C) 20 cm (D) 50 cm

Tardigrade Admin · Accepted Answer

Let the power of lenses are P1 and P2 respectively. When they are separated by distance d, the equivalent power of the combination, Peq = P1 + P2 - 2dP1P2 = 10 + 10 - 2d (10× 10) = 20 - 100 d Now, 20 - 100 d < 0 ⇒ 100d < 20 d > (1/5) m ⇒ d > (100/5) cm or 20 cm

Q. Two lenses have 10D power each and they are separated by a distance. Beyond which distance does the power of combination changes from positive to negative?

5 cm

10 cm

20 cm

50 cm

Let the power of lenses are P1​ and P2​ respectively. When they are separated by distance d, the equivalent power of the combination, Peq​=P1​+P2​−2dP1​P2​ =10+10−2d(10×10) =20−100d Now, 20−100d<0 ⇒100d<20 d>51​m ⇒d>5100​cm or 20cm

Q. Two lenses have $10 D$ power each and they are separated by a distance. Beyond which distance does the power of combination changes from positive to negative?