
The far point of a myopia eye is 1.5m. To correct this defect of the eye, the power of the lens is ________.
A.)0.66D
B.)-0.66D
C.)+1.55D
D.)-1.55D
Answer
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Hint- In order to solve these types of questions, we will use the concept of power of the lens and we will proceed further first by calculating the focal length using the lens formula and then proceed further for finding the power of the lens.
Formula used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where f is the focal length, v is the image distance and u is the object distance.
Complete step-by-step answer:
Let f be the focal length, v be the image distance and u be the object distance.
As we know that the for an object at infinity
Object distance $u = - \infty $
Image distance is given
$v = - 1.5m$
Now we will calculate the focal length using the lens formula
The lens formula is given as
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Substituting the value of u and v in the above formula
$\dfrac{1}{f} = \dfrac{1}{{ - 150}} - \dfrac{1}{{ - \infty }}$
$f = - 1.5m$
Now we will calculate the power of lens
The power of lens is given by
\[P = \dfrac{1}{f}\]
Substituting the value of f in the above formula
$
P = \dfrac{1}{{ - 1.5}} \\
P = - 0.67D \\
$
The power of lens is -0.67D
Hence, the correct option is B.
Additional Information- Lenses can be classified into two major forms of large measure: convex and concave. The lenses in the middle are smoother than the corners, while the lenses along the sides are concave. The mirror concentrates on a spot on the other side of the frame, a light ray traveling through a convex prism. This is considered the focus of publicity. In the case of concave lenses that diverge from the condensation of light waves, the focal point is on the source of incoming light that appears to come from the direction of light through the lens.
Note- In order to solve these types of questions, apart from the formulas you must remember the sign conventions. Some of the sign conventions are- for convex lens the focal length is positive and for concave lens the focal length is negative. Similarly others remember the other conventions. Most optical devices make use of not just one lens, but of a combination of convex and concave lenses.
Formula used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where f is the focal length, v is the image distance and u is the object distance.
Complete step-by-step answer:
Let f be the focal length, v be the image distance and u be the object distance.
As we know that the for an object at infinity
Object distance $u = - \infty $
Image distance is given
$v = - 1.5m$
Now we will calculate the focal length using the lens formula
The lens formula is given as
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Substituting the value of u and v in the above formula
$\dfrac{1}{f} = \dfrac{1}{{ - 150}} - \dfrac{1}{{ - \infty }}$
$f = - 1.5m$
Now we will calculate the power of lens
The power of lens is given by
\[P = \dfrac{1}{f}\]
Substituting the value of f in the above formula
$
P = \dfrac{1}{{ - 1.5}} \\
P = - 0.67D \\
$
The power of lens is -0.67D
Hence, the correct option is B.
Additional Information- Lenses can be classified into two major forms of large measure: convex and concave. The lenses in the middle are smoother than the corners, while the lenses along the sides are concave. The mirror concentrates on a spot on the other side of the frame, a light ray traveling through a convex prism. This is considered the focus of publicity. In the case of concave lenses that diverge from the condensation of light waves, the focal point is on the source of incoming light that appears to come from the direction of light through the lens.
Note- In order to solve these types of questions, apart from the formulas you must remember the sign conventions. Some of the sign conventions are- for convex lens the focal length is positive and for concave lens the focal length is negative. Similarly others remember the other conventions. Most optical devices make use of not just one lens, but of a combination of convex and concave lenses.
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