Answer:
The power will be "2.08 D".
Explanation:
According to the question,
The location of the image form lens,
[tex]d_i=50-2[/tex]
[tex]=48 \ cm[/tex]
We know that,
⇒ [tex]\frac{1}{d_i} -\frac{1}{d_o} =\frac{1}{f}[/tex]
here,
[tex]d_o[/tex] = Object location
[tex]d_i[/tex] = Image location
[tex]f[/tex] = Focal length
On putting the values in the above expression, we get
⇒ [tex]\frac{1}{48} -\frac{1}{\infty} =\frac{1}{f}[/tex]
[tex]f=48 \ cm[/tex]
or,
[tex]f=0.48 \ m[/tex]
hence,
The refractive power will be:
⇒ [tex]P=\frac{1}{f}[/tex]
[tex]=\frac{1}{0.48}[/tex]
[tex]=2.08 \ D[/tex]
