Single Choice

To remove myopia ( short sightedness ) a lens of power $$0.66\ D$$ is required. The distant point of the eye is approximately

A$$100\ cm$$
B$$150\ cm$$
Correct Answer
C$$50\ cm$$
D$$25\ cm$$

Solution

ATQ,
Far point of the eye = focal length of the lens
$$=\dfrac{100}{P}=\dfrac{100}{0.66}=151\ cm$$

SIMILAR QUESTIONS

Optics

Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

Optics

A defective eye cannot see close objects clearly because their image is formed:

Optics

Identify the wrong description of the given figures.

Optics

For the myopic eye, the defect is cured by

Optics

Match the list I with the List II for the combination shown Presbiopia Sphero-cylindrical lens Hypermetropia Convex lens of proper power may be used close to the eye Astigmatism Concave lens of suitable focal length Myopia Bifocal lens of suitable focal length

Optics

A person is suffering from 'presbyopia' ( myopia and hyper metropia both defects ) should use

Optics

A student sitting at the back of the classroom cannot read clearly the letters written on the blackboard. What advice will a doctor give to her? Draw ray diagram for the correction of this defect.

Optics

When do we consider a person to be myopic or hypermetropic? Explain using diagrams how the defects associated with myopic and hypermetropic eye can be corrected?

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