Answer:
Part A
The nature of the lens required is a concave lens
Part B
The power of the lens required is -1.25 D
Explanation:
Part A
Myopia is the eye condition whereby the focus of light entering the eye is in front of the retina such that the vision of a myopic person is blurred
The given far point of the myopic person in question = 80 cm
To correct the problem, the focus of light entering the eye will be moved further to reach the retina, by means of a concave lens which makes distant objects appear to come from (form an image at) 80 cm
Part B
The lens formula is given as follows;
[tex]\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}[/tex]
Where;
f = The focus of the lens (virtual focus)
[tex]d_o[/tex] = The distance of the object = ∞ (infinity)
[tex]d_i[/tex] = The distance of the image = -80 cm (virtual image formed in front of the mirror)
Plugging in the values, gives;
[tex]\dfrac{1}{f} = \dfrac{1}{\infty} + \dfrac{1}{-80 \, cm} = \dfrac{1}{-80 \, cm}[/tex]
[tex]\therefore \dfrac{1}{f} = \dfrac{1}{-80 \, cm}[/tex]
f = -80 cm = -0.8 m
The power of a lens, 'P' is the reciprocal of the focal length, 'f', given in meters
[tex]P = \dfrac{1}{f (in \, meters)}[/tex]
The SI unit for the power of a lens is the dioptre (D)
Therefore;
[tex]P = \dfrac{1}{-0.80 \, m} = -1.25 \, D[/tex]
Therefore, the power of the lens required, P = -1.25 D.
