An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.

Explanation

For convex lens,
Object distance u = - 40 c, focal length
f = + 30 cm
Lens formula ,
`(1)/(f_1) = (1)/(v_1) - (1)/(u_1)`
`therefore (1)/(v_1) = (1)/(f_1) + (1)/(u_1)`
`therefore (1)/(v_1) = (1)/(30) - (1)/(40) = (4-3)/(120) = (1)/(120)`
`therefore v_1 = 120 cm `
`rArr` Magnification by 1st (convex) lens,
`m_1 = (v)/(|u|) = 120/40 = 3 `
`rArr` This image is virtual for another lens,
`therefore u_2 = (120 -8) = + 112 cm `
`f_2 = - 20 cm `
From lens formula ,
`(1)/(f_2) = (1)/(v_2) - (1)/(u_2)`
`therefore (1)/(v_2) = (1)/(-20) + (1)/(112) = (-112 + 20)/(112 xx 20) = (-92)/(112 xx 20)`
`therefore (1)/(v_2) = - (112 xx 20)/(92) = - 24. 9 cm `
`rArr` Magnification by another lens (concave),
`m_2 = (|v_2|)/(u_2) = (112 xx 20)/(92 xx 112) = (20)/(92) ~~ 0.217 `
`rArr` Resultant magnification of combination,
`m = m_1 xx m_2`
`= 3 xx 0.217`
= 0.651
`rArr` Image height
`m = (h_2)/(h_1)`
`therefore h_2 = m xx h_1 = 0.651 xx 1.5`
`therefore h_2 = 0.9765 `
`therefore h_2 = 0.9765`
`therefore h_2 ~~ 0.98` cm