Consider the situation shown in figure. The refractive index of medium is `3//2` and its radius of curvature is 330 cm `PO_1=PO_2=10cm` Find distance `O_1` and `O_2` and seen by (a) `O_1` and (b) `O_2`

Explanation

(a). `O_1` is observing `O_2`:
`mu=(3)/(2),mu_2=1.0,mu=-10cm,R=30cm,v=?`
`(mu_2)/(v)-(mu_1)/(u)=(mu_2-mu_1)/(R)`
`(1)/(v)-((3)/(2))/(-10)=(1-3//2)/(30)implies(1)/(v)+(3)/(20)=(1)/(60)`
`(1)/(v)=(1)/(60)-(3)/(20)=(1-9)/(60)`
`v=-(60)/(8)=-7.5cm`, image of `O_(2)` is formed at `I`
`O_(1)l=O_(1)PI=10+7.5=17.5cm`
(b). `O_(2)` is observing `O_(1)`
`mu_1=1,mu_2=(3)/(2),u=-10cm,R=-30cm,v=?`
`(mu_2)/(v)-(mu_1)/(u)=(mu_2-mu_1)/(R)implies(3//2)/(v)-(1)/(-10)=(3//2-1)/(-30)`
`(3)/(2v)=-(1)/(60)-(1)/(10)=(-7)/(60)`
`v=(-60)/(7)cm`, image of `O_(1)` is formed at `I`
`PI(60)/(7)cm,O_(2)I=O_(2)P+P+PI=10+(60)/(7)=(130)/(7)cm`