Question

3- state the mi se 409. formula for determining
the focal length of sphenicki mi y 484 White
the meaning of the symbols used an object
is placed at a distance of ascm - from a concave
minston of focal length 15cń calculate the distance
of the image from the mirror .​

Answer 1

Explanation:

Hypermetropia can be corrected by using a convex lens. A convex lens converges the incoming light such that the image is formed on the retina.

An object at 25 cm forms an image at the near point of the hypermetropic eye. Here, near point is 1 m.

Given,

Object distance,u=−25 cm

Image distance, v=−100 cm

From lens formula,

v

1

−

u

1

=

f

1. 1

−100

1

−

−25

1

=

f

1

Focal length,f=100/3 cm=1/3 m

Power, P=

f

1

=

1/3

1

=3 D

solution

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Answer 2

Explanation:

Hypermetropia can be corrected by convex lens. A convex lens converges the incoming light such that the image is formed on the retina. An object at 25 cm forms an image at the near point of the hypermetropic eye. Here, near point is 1 m. Given, Object distance,u=-25 cm Image distance, v=-100 cm 1. From lens formula, V 1