Question

A person suffering from myopia (near-sightedness) was advised to
wear corrective lens of power - 2.5 D. A spherical lens of same
focal length was taken in the laboratory. At what distance should a
student place an object from this lens so that it forms an image at
a distance of 10 cm from the lens ?



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Answer 1

Answer:

• Position of object should be 8 cm infront of the Lens.

Explanation:

Person suffering from Myopia was advised to wear corrective lens of power -2.5 D

since, Power = 1 / focal length

therefore,

→Focal length of lens = 1 / (-2.5)

→Focal length of lens = -0.4  m

Converting it to cm

→ Focal length of lens, f = -40 cm

And, Its a concave lens used for correction of Myopia.

Now, this lens was taken to laboratory

Since, Image formed by Concave lens is always virtual and erect therefore,

→ Position of image, v = -10 cm

We need to find position of object, u = ?

Using Lens formula

→ 1/f = 1/v - 1/u

→ 1/(-40) = 1/(-10) - 1/u

→ 1/u = (1/-10) - (1/40)

→ 1/v = (1/-10) - (1/40)

→ 1/v = ( - 4 - 1 ) / 40

→ 1/v = -5 / 40

→ v = 40/-5

→ v = -8 cm

Therefore,

• Student must put the object at 8 cm distance infront of the Lens.

Answer 2

Explanation:

Nature of Spherical lens = Convex

Given :-

u = - 45 cm

v = + 90 cm

h₁ = + 2 cm

Solution :-

Using lens formula, we get

1/f = 1/v - 1/u

⇒ 1/f = 1/90 - 1/- 45

⇒ 1/f = 1/90 + 1/45

⇒ 1/f = 1 + 2/90

⇒ 1/f = 3/90

⇒ 1/f = 1/30

⇒ f = + 30 cm

Again, m = h₂/h₁ = v/u

h₂ = v/u × h₁

⇒ h₂ = 90/- 45 × 2

⇒ h₂ = - 4 cm

Hence, the height of image is - 4 cm. Negative sign indicates that it is formed below the principal axis.