A student can distinctly see the object upto a distance 15 cm. He wants to see the black board at a distance of 3 m. Focal length and power of lens used respectively will be :

Step-by-step solution

The student should use a lens which forms image at distance of 15 cm of the object placed at 3 m i.e., object distance u=-3 ~m =-300 ~cm , image distance v=-15 ~cm From lens formula 1 v - 1 u = 1 f we get 1 -15 - 1 -300 = 1 f or 1 f = 1 300 - 1 15 =- 19 300 or f= -300 19 =-15.8 ~cm (concave lens) Now power of the lens is P= 100 f( ~cm ) = 100 -15.8 =-6.33 D