1.

As shown in figure, a plano-concave lens is placed in such a way that it becomes completely fit with plano-convex lens. Their plane surfaces are parallel. If their refractive indices are 1.6 and 1.5 respectively and radius of curvature is R, then focal length of combination is .....

Answer»

`(R)/(6.2)`
`(R)/(3.1)`
`(R)/(0.2)`
`(R)/(0.1)`

Solution :`f=(R)/(mu_1-mu_2)=(R)/(1.6-1.5)=(R)/(0.1)`
For first LENS,` (1)/(f_1)=(n-1)((1)/(R_1)-(1)/(R_2))`
`R_1= INFTY,R_2=-R,n=mu_1`
`(1)/(f_1)=(mu_1-1)((1)/(infty)+(1)/(R))`
`therefore (1)/(f_1)=(mu-1)/(R)` … (1)
For SECOND lens, `(1)/(f_2)=(n-1)((1)/(R_1)-(1)/(R_2))`
`R_1=-R,R_2=infty,n-mu_2`
`(1)/(f_2)=(mu_2-1)(-(1)/(R)-(1)/(infty))`
`infty (1)/(f_2)=(1-mu_2)/(R)` ... (2)
Focal LENGTH of combination,`1/f=(1)/(f_1)+(1)/(f_2)=(mu_1-1)/(R)+(1-mu_2)/(R)`
`therefore 1/f=(mu_1-mu_2)/(R)`
`therefore f=(R)/(mu_1-mu_2)=(R)/(1.6-1.5)=(R)/(0-1)`


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