Question

a man without his eye glasses can read a book holding it 13.9 cm far from his eyes what is the optical power of the lens of it is eye glasses so that he could read a book at the distant of best vision​

Answer 1

Answer:

7.19 D

Explanation:

In this question,

• Focal length (f) = 13.9 cm

Apply power formula and find out the Optical power of the lens required.

$\boldsymbol{Power=\dfrac{1}{Focal \: length\: \: (in \: metres)}}$

$\implies \sf{Power = \dfrac{100}{Focal\: length\:(in \: cm)}}$

$\implies \sf{Power = \dfrac{100}{13.9}}$

$\implies \sf{Power= 7.19\:D}$

$\therefore \boxed{\bf{Power \: of \: lens\:required = 7.19\:D}}$

KNOW MORE:

• Power is defined as the degree of convergence of a lens.
• It is also the inverse of focal length.
• The SI unit of Power is Dioptre (D).

Answer 2

Answer:

Answer:

7.19 D

\begin{gathered}\\\end{gathered}

Explanation:

In this question,

Focal length (f) = 13.9 cm

Apply power formula and find out the Optical power of the lens required.

\begin{gathered}\boldsymbol{Power=\dfrac{1}{Focal \: length\: \: (in \: metres)}}\\\\\end{gathered}

Power=

Focallength(inmetres)

1

\begin{gathered}\implies \sf{Power = \dfrac{100}{Focal\: length\:(in \: cm)}}\\\\\end{gathered}

⟹Power=

Focallength(incm)

100

\begin{gathered}\implies \sf{Power = \dfrac{100}{13.9}}\\\\\end{gathered}

⟹Power=

13.9

100

\begin{gathered}\implies \sf{Power= 7.19\:D}\\\\\end{gathered}

⟹Power=7.19D

\begin{gathered}\therefore \boxed{\bf{Power \: of \: lens\:required = 7.19\:D}}\\\\\end{gathered}

∴

Poweroflensrequired=7.19D