Question

What is the focal length having power of +4.5D

Answer 1

Answer:

Given:

Power of a lens = +4.5D.

To find:

Focal length of lens.

Concept:

Power of lens is an index of understanding the converging or diverging capability of a lens.

If power is positive , then it's a Convex lens . If power is negative, then it's a Concave lens.

The reciprocal of power gives focal length of lens in metres unit.

$\: \: \: \: \: \: \: \: \boxed{ \large{ \sf{ \red{ \bold{f(in \: m) = \dfrac{1}{power}}}}}}$

Calculation:

$\therefore \: f = \dfrac{1}{p}$

$= > f = \dfrac{1}{4.5}$

$= > f = 0.2222 \: metres$

$= > f = 22.22 \: cm$

So final answer is :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \large{ \sf{ \bold{ \blue{ f = 22.22 \: cm}}}}}$

Additional information:

Since power or Focal length of the given lens is coming positive ,we can understand that :

• Lens is Convex lens
• Lens is converging in nature.

Answer 2

$\underline{ \huge{ \huge{ \boxed{ \bold{ \sf{ \orange{Answer}}}}}}} \\ \\ \star \sf \: \bold{ \red{Given}} \\ \\ \mapsto \sf \: power \: of \: lens = + 4.5 \: D \\ \\ \star \sf \: \bold{ \red{To \: Find}} \\ \\ \mapsto \sf \: focal \: length \: of \: lens \\ \\ \star \sf \: \bold{ \red{Formula}} \\ \\ \mapsto \sf \: relation \: between \: power \: of \: lens \: and \: \\ \sf \: focal \: length \: is \: given \: as \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\boxed{ \bold{ \sf{ \huge{ \purple{P = \frac{1}{f} }}}}}} \\ \\ \mapsto \sf \: here \: unit \: of \: f \: is \: m \\ \\ \star \sf \: \bold{ \red{Calculation}} \\ \\ \mapsto \sf \: f = \frac{1}{P} = \frac{1}{4.5} = 0.2222 \: m \\ \\ \: \: \: \: \: \: \: \: \: \: \:\star \: \: \underline{\boxed{ \sf{ \bold{ \blue{f = 22.22 \: cm}}}}} \: \: \star \\ \\ \mapsto \sf \: here \: power \: is \: positive \: so \: lens \: is \: Convex.$