The focal length of the lens to be used to correct hypermetropia when the near point distance is “d” cm. 

Correct option is  C) 2f

Correct option is  B) double

Correct option is  A) -37.5 cm

Correct option is  C) 30 cm

Correct option is  A) 20 cm

Correct option is  C) 40 cm

Correct option is  B) f = -d

Correct option is  A) 2.27 cm

 U = -12 f = -10  1/v - 1/u = 1/f 1/v + 1/12 = 1/10 1/v = 1/60  \(-\frac{1}{v^2} \)dv or \(\frac{1}{u^2} \)du du = \(\frac{v^2}{u^2} \)du \( \frac{v}{u} \)= same path △u1 = △u2 = 1mm △u1 = △v2  △v2 = vf - vi  x2 = △v2 + 1mm x2...