Answer:

(a) To find the distance of the boy's retina from the eye-lens, we can use the lens formula:

1/f = 1/v - 1/u,

where f is the focal length of the lens, v is the image distance, and u is the object distance.

Given that the boy's eye-lens can adjust power between 50 D and 60 D, we know that the focal length (f) of the lens can vary between 1/f = 50 cm^-1 and 1/f = 60 cm^-1.

Since the far point is considered to be at infinity, the object distance (u) can be taken as infinite, resulting in 1/u = 0. Therefore, the lens formula simplifies to:

1/f = 1/v.

Taking the maximum focal length of 60 cm^-1, we can substitute it into the lens formula:

1/60 = 1/v.

Solving for v, we find:

v = 60 cm.

Hence, the distance of the boy's retina from the eye-lens is 60 cm.

(b) The near point is the closest distance at which the boy can see objects clearly without strain. In general, the near point is determined by the maximum power of accommodation of the eye, which is the reciprocal of the lens power. In this case, the maximum lens power is 60 D.

To find the near point, we use the formula:

Near point = 100 cm / maximum lens power.

Substituting the maximum lens power of 60 D into the formula:

Near point = 100 cm / 60 D = 1.67 cm^-1.

Converting this to centimeters:

Near point = 1 / 1.67 cm^-1 = 0.6 cm.

Therefore, the boy's near point is approximately 0.6 cm, or 10 cm as mentioned in the question.

Explanation: The near point is determined by the maximum lens power of the eye, which represents the maximum ability of the lens to adjust and focus on nearby objects. In this case, with a maximum lens power of 60 D, the boy can focus on objects as close as 10 cm away, which becomes his near point.