Question

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A simple question for you all.

Two thin lenses of power +2.5D and- 0.5D are put in contact, what is the power, focal length of the combination?
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Answer 1

Answer:

• The Power and focal length are 2.0 D & 0.5 m respectively.

Explanation:

Given, Two lenses have power of +2.5 D and - 0.5 D and they are put in close contact.

From the formula we know that,

⇒ P = P₁ + P₂

(Here you need to add them with sign conventions)

Substituting the values,

⇒ P = + 2.5 + (- 0.5 )

⇒ P = + 2.5 - 0.5

⇒ P = 2.0 D

∴ The Power of the combined lens is 2.0 D.

Now, for calculating focal length,

⇒ f = 1/P

Substituting the values,

⇒ f = 1/2

⇒ f = 0.5 m

⇒ f = 0.5 m (or) 50 cm

(Here f is positive which indicates that the combined effect of lens will have converging effect )

∴ The focal length of combined lens is 0.5 m.

Answer 2

Question :-

Two thin lenses of power +2.5D and -0.5D are put in contact, what is the power, focal length of the combination?

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Solution :-

➲ Given Information :-

• Power of Lens 1 ( Power₁ ) ➢ +2.5 D
• Power of Lens 2 ( Power₂ ) ➢ -0.5 D

➲ To Find :-

• Power of the combination ( Power₁ + Power₂ )
• Focal Length of the combination

➲ Concept :-

• Simple numericals based on Refraction of Light through Lenses

➲ Formulae Used :-

1. $\boxed{\boxed{\bf{\red{Power = Power_1 + Power_2}}}}$
2. $\boxed{\boxed{\bf{\green{Focal\: Length = \dfrac{1}{Power}}}}}$

➲ Explanation :-

• In order to find the power of the two lenses, We will add the power of each lens. The resultant will then be put in the formula mentioned above, used to calculate Focal Length. After some minute calculations, we'll have our answers, i.e., Power of the combination and Focal Length of the combination of lenses. So let's proceed towards our calculation.

➲ Calculation :-

• Firstly Calculating the Power of the combination.

Substituting the values given in the first formula, We get,

$\boxed{\boxed{\rm{\red{Power = Power_1 + Power_2}}}}$

Substituting the values, We get,

$\rm:\implies{\purple{Power = +2.5 \:D + ( - 0.5 \:D )}}$

$\rm:\implies{\purple{Power = +2.5\: D - 0.5\: D }}$

$\rm:\implies{\purple{Power = \bf{+2.0 \:D}}}$

∴ The Power of the combination of lenses is 2.0 D

• Now calculating the Focal Length of the combination.

Substituting the values given in the second formula, We get,

$\boxed{\boxed{\rm{\green{Focal\: Length = \dfrac{1}{Power}}}}}$

Substituting the values, We get,

$:\implies\rm{\blue{Focal\: Length = \dfrac{1}{2.0}}}$

$\because\:(\rm{\dfrac{1}{2.0}} \: can \:also \:be \:written \:as\: 0.5)$

$\rm:\implies{\blue{Focal\: Length = \dfrac{1}{2.0} = 0.5 \:m \: or\: 50\: cm}}$

∴ The Focal Length of the combination of lenses is 0.5 m or 50 cm

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Final Answer :-

1. The Power of the combination of lenses is 2.0 D
2. The Focal Length of the combination of lenses is 0.5 m or 50 cm

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