st lens:Object distance from the lens i.e. u is -40 cm.Let us assume that the image formation of the first lens takes place at the SCREEN which is 'X' cm AWAY.Using lens formula,1/F = 1/v - 1/uSubstitute the values,1/f = 1/x - 1/(-40)1/f = 1/x + 1/401/f = (40 + x)/40xf = 40x/(40 + x)40f + fx = 40x40f = 40x - fx40f = x(40 - f)40f/(40 - f) = x ...............(1st equation)For second lens:Object distance i.e. u from the screen is -80 cm and again the image formation i.e. v is 'x - 40' cm.Since the first lens image formation was x cm always from the lens and for the second ONE object is 80 cm away. So, object will be place at image distance from lens - object distance from the first lens. i.e. (x - 40)Using lens formula,1/f = 1/v - 1/uSubstitute the values,1/f = 1/(x-40) - 1/(-80)1/f = 1/(x-40) + 1/80 ...............(2nd equation)Substitute value of (1st equation) in (2nd equation)→ 1/f = 1/[40f/(40-f) - 40] + 1/80→ 1/f = 1/[(40f - 1600 + 40f)/(40 - f)] + 1/80→ 1/f = 1/[(80f -1600)/(40 - f)] + 1/80→ 1/f = (40 - f)/(80f -1600) + 1/80→ 1/f = 1/80 × (40 - f)/(f - 20) + 1/1→ 1/f = 1/80 × (40 - f + f - 20)/(f - 20)→ 1/f = 1/80 × (20)/(f - 20)→ 1/f = 20/80 × 1/(f - 20)→ 1/f = 1/4 × 1/(f - 20)→ 4/f = 1/(f - 20)Cross-multiply them,→ 4f - 80 = f→ 4f - f = 80→ 3f = 80→ f = 80/3Therefore, the focal length of the lens is 80/3 cm.Option (4) 80/3 cm