Determine the .effective focal length. of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all .

Explanation

(i) Here `f_1 = 30 cm , f_2 = -20 cm , d= 8.0 cm`
Where d = distance between two lenses.
`rArr` If parallel rays incident on first lens in absence of second lens,
`u_1 = oo ` and `f_1 = 30 cm`
`therefore` From lens formula
`1/f_1 = 1/v_1 - 1/u_1`
`therefore 1/30 =1/v_1 - 1/(-oo) = 1/(v_1) - 0`
`therefore 1/30 = 1/v_1`
`therefore v_1 = 30 cm `
This image wil work as virtual object for second lens.
`therefore u_e = (30 -8) = + 22 cm f_2 = - 20 cm `
From lens formula
`1/f_2 = 1/v_2 - 1/u_2`
`therefore 1/v_2 = 1/(-20) + (1)/(22) = (-22 + 20)/(440) = (-2)/(440)`
`therefore v_2 = - (440)/(2) = - 220 cm`
`rArr` Parallel rays coverged at 220-4 = 216 cm from mid-point of system two lenses.
(ii) Let, parallel beam first incidents at left to concave mirror,
`therefore u_1 = - oo ,f_1 = -20` cm
From lens formual,
`1 /f_1 = 1/v_1 - 1/u_1`
`therefore 1/v_1 = 1/f_1 + 1/u_1`
`therefore 1/v_1 = (1)/(-20) + 1/(-oo) = - (1)/(20) + 0`
`therefore v_1 = -20 cm `
`rArr` This image wil work as real image for second lens,
`therefore u_1 = - (20 + 8) = -28 cm `
`f_2 = 30 cm `
From lens formula
`(1)/(f_2) = (1)/(u_2) - (1)/(u_2)`
`therefore (1)/(v_2) = (1)/(f_2) + (1)/(u_2)`
`therefore (1)/(v_2) = (1)/(3) + (1)/(-28) = (14 -15)/(420) = - (1)/(420)`
`therefore v_2 = - 420 cm `
The parallel beam appears to diverge from a point 420 - 4 = 416 cm on the left of the centre of the two lens system. From the above tow cases it concludes that the answer depends on which side of the lens system the parallel beam in incident. So, the notion of effective focal lenght doesn.t seem to ve useful here.