A myopic adult has a far point at 0.1 m. His power of accommodation is 4D.

Statement 1:  Power of lens are required to see distant object is $-10D$

Statement 2 : Nearest point without glasses is $0.07m$

i) If for the normal relaxed eye of an average person, the power at the far point be $P_f$. The required power $P_f=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.2}=60D$

By the corrective lens the object distance at the far point is $\infty .$

The power required is, $P_f\text{'}=\frac{1}{f\text{'}}=\frac{1}{\infty }+\frac{1}{0.02}=50D$

Now for eye + lens system, we have the sum of the eye and that of the glasses $P_g$

$P{\text{'}}_f=P_f+P_g\Rightarrow 50D=60D+P_g$

Which gives, $P_g=-10D$

ii) For the normal eye his power of accommodation is 4D. Let the power of the normal eye for near vision be $P_n.$Then, $4=P_n-P_{f}or{P}_n=64D$

Let his near point be $x_n,$ then $\frac{1}{x_n}+\frac{1}{0.02}=64or\frac{1}{x_n}+50=64$

$\frac{1}{x_n}=14\Rightarrow x_n=\frac{1}{14}m=0.07m$