एक उत्तल लेंस की हवा में फोकस-दुरी 20 cm है। बताएँ की पानी में डुबाने पर इसकी फोकस-दुरी क्या होगी? (हवा-काँच युग्म का अपवर्तनांक 1.5 और हवा-पानी युग्म का अपवर्तनांक 1.33 है।)

Explanation

हम जानते हैं कि-
`(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))`
हवा में,
`(1)/(f_(a)) = (._(a) mu_(g)) ((1)/(R_(1)) - (1)/(R_(2)))" "...(i)`
पानी में,
`(1)/(f_(w)) = (._(w) mu_(g)) ((1)/(R_(1)) - (1)/(R_(2)))" "...(ii)`
समी. (i) को (ii) से भाग देने पर,
`(f_(w))/(f_(a))= (._(a) mu_(g) - 1)/(._(w) mu_(g) - 1)" "...(iii)`
हम जानते हैं कि
`._(a) mu_(omega) xx ._(w)mu_(g) xx ._(g) mu_(a) = 1`
`rArr" "._(w) mu_(g) = (1)/(._(a) mu_(w) xx ._(g) mu_(a)) = (._(a)mu_(g))/(._(a) mu_(w))" "...(iv)`
`(f_(w))/(f_(a)) = (._(a) mu_(g) - 1)/((._(a)mu_(g))/(._(a) mu_(w)) - 1) = ((._(a) mu_(g) - 1) ._(a) mu_(w))/(._(a) mu_(g) - ._(a) mu_(w))`
`= (._(a) mu_(g) xx ._(a) mu_(w) - ._(a) mu_(w))/(._(a) mu_(g) - ._(a) mu_(w))`
यहाँ `._(a) mu_(g) = (3)/(2), ._(a) mu_(2) = (4)/(3)`
`therefore" "(f_(w))/(f_(a)) = ((3)/(2) xx (4)/(3) - (4)/(3))/((3)/(2) - (4)/(3)) = 4`
`therefore" "f_(w) = 4f_(a) = 4 xx 20 = 80` cm