The ciliary muscles of eye control the curvature of the lens in the eye and hence and alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means, radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 25 cm for a grown up person.
A person can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25 cm.
A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image of the spectacle lens becomes object for the eye lens and whose image is formed on the retina.
The number of spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100/3 cm), then number of lens will be +3.
For all the calculations required, you can used the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens.
A far sighted man cannot see objects clearly unless they are at least 100 cm from his eyes. The number of the spectacle lenses that will make his range of clear vision equal to an average grown up person will be
The ciliary muscles of eye control the curvature of the lens in the eye and hence and alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means, radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 25 cm for a grown up person.
A person can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25 cm.
A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image of the spectacle lens becomes object for the eye lens and whose image is formed on the retina.
The number of spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100/3 cm), then number of lens will be +3.
For all the calculations required, you can used the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens.
A far sighted man cannot see objects clearly unless they are at least 100 cm from his eyes. The number of the spectacle lenses that will make his range of clear vision equal to an average grown up person will be
The correct option is C +3
For the far-sighted man, lens should make image of the nearby object at distances beyond 100 cm. For grown up person least distance is 25 cm. For lens, u=−25,v=−100
1f=1−100−1(−25)⇒1f=3100
P=+3 so no. of spectacle is =+3.
For the far-sighted man, lens should make image of the nearby object at distances beyond 100 cm. For grown up person least distance is 25 cm. For lens, u=−25,v=−100
1f=1−100−1(−25)⇒1f=3100
P=+3 so no. of spectacle is =+3.
