Question

Two lenses have power of + 2D and - 4D. What is the nature and focal length of each lens.? an object is kept at a distance of 100 cm in front of each of the above lenses calculate 1 the image distance in each. 2 magnification in each

Answer 1

The image distance of first and second lens is 100 cm and -20 cm respectively. The value of magnification for convex lens  is 1 and for concave lens is 0.2.

Explanation:

Two lenses are given having power of +2D and - 4D. The power of a convex lens is positive while the power of a concave lens is negative.

Focal length of lens 1,

$f_1=\dfrac{1}{2}=0.5\ m=50\ cm$

Focal length of lens 2,

$f_2=\dfrac{1}{-4}=0.25\ m=25\ cm$

For lens 1,

Object distance, u = -100 cm

Let v is the image distance.

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{50}+\dfrac{1}{(-100)}$

v = 100 cm

Magnification,

$m=\dfrac{v}{u}=\dfrac{100}{-100}=-1$

For lens 2,

Object distance, u = -100 cm

Let v is the image distance.

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-25}+\dfrac{1}{(-100)}$

v = -20 cm

Magnification,

$m=\dfrac{v}{u}=\dfrac{-20}{-100}=0.2$

Hence, this is the required solution

Lens formula

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