1.

. A person needs a lens pf of power -5.5D for correcting his distant vision. For correcting his nearvision, he needs a lens of power +1.5D. Calculate the focal length of the lenses used in these two cases.​

Answer»

Given:-

  • Power of LENS ,P₁ = -5.5D

  • Power of lens , P₂ = 1 .5 D

To Find:-

  • Focal length ,f in both case.

Solution:-

As we KNOW that Power of lens is defined as the reciprocal of its focal length

→ P = 1/f

CALCULATING focal length of 1st lens

→ P₁ = 1/f₁

Substitute the VALUE we get

→ -5.5D = 1/f

→ -5.5 ×f = 1

→ f₁ = 1/(-5.5)

→ f₁ = -0.18 m

∴ The focal length of 1st lens is 0.18m.

Now Calculating focal length of 2nd lens

→ P₂ = 1/f₂

Substitute the value we get

→ 1.5 = 1/f₂

→ 1.5×f₂ = 1

→ f₂ = 1/1.5

→ f₂ = 0.06 m

∴ The focal length of 2nd lens is 0.06 metre.


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