
(a) Define the power of the lens.
(b) Draw a ray diagram for formation of image of an object situated at a point in between $2{{f}_{1}}$ and ${{f}_{1}}$ distances from the optical centre of the concave lens and write the nature of the image formed.
(c) What happens when light is falling perpendicular to an interface of two media?
Answer
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Hint: Recall the definition of the power of lens and hence answer the first part. Keeping all the points used for the image formation of the concave lens in mind, form the image of the object kept between ${{f}_{1}}$ and $2{{f}_{1}}$. Recall Fermat’s principle and then answer the c-part of the question.
Complete answer:
(a) In the first part of the question we are asked to define the power of the lens.
We define power as the reciprocal of focal length. But the focal length in the expression should be in meters. So we could say that lenses with short focal lengths are more powerful. The unit of power of the lens is dioptres (D).
Also, for converging lenses, (that is, convex lenses) the focal lengths are positive which further results in positive power of the lens. And for diverging lenses, (that is, concave lenses) the focal lengths are negative, hence the power will also be negative. The power (P) of the lens is given by,
$P=\dfrac{1}{f}$
Where, f is in meters.
So, 1 dioptre is the power of a lens whose focal length is 1 meter.
(b) We are asked to draw the ray diagram of image formation of an object kept in between $2{{f}_{1}}$ and${{f}_{1}}$. We know that concave lenses are diverging lenses so they diverge the ray falling on it.
We could form the image by using these two points:
(1) A light ray parallel to the principal axis diverges in such a way that it appears to come from the focal point (first focal point in this case).
(2) A light ray passing through the optical centre goes straight (that is, doesn’t deviate).
So, we see that the image is located on the object side of the lens. Also the image formed is virtual but upright and is also diminished in size when compared to the object.
(c) When light falls perpendicular to an interface of two media, it goes straight forward without bending. This could be explained by Fermat’s principle that states that the light will take that path that uses the least time.
Note:
Image formation is quite simple if you keep some important points in mind. Other than the two rules used in the solution, we also have another rule. This rule states that the ray going towards focus on another side becomes parallel to the principal axis after refraction.
Complete answer:
(a) In the first part of the question we are asked to define the power of the lens.
We define power as the reciprocal of focal length. But the focal length in the expression should be in meters. So we could say that lenses with short focal lengths are more powerful. The unit of power of the lens is dioptres (D).
Also, for converging lenses, (that is, convex lenses) the focal lengths are positive which further results in positive power of the lens. And for diverging lenses, (that is, concave lenses) the focal lengths are negative, hence the power will also be negative. The power (P) of the lens is given by,
$P=\dfrac{1}{f}$
Where, f is in meters.
So, 1 dioptre is the power of a lens whose focal length is 1 meter.
(b) We are asked to draw the ray diagram of image formation of an object kept in between $2{{f}_{1}}$ and${{f}_{1}}$. We know that concave lenses are diverging lenses so they diverge the ray falling on it.
We could form the image by using these two points:
(1) A light ray parallel to the principal axis diverges in such a way that it appears to come from the focal point (first focal point in this case).
(2) A light ray passing through the optical centre goes straight (that is, doesn’t deviate).
So, we see that the image is located on the object side of the lens. Also the image formed is virtual but upright and is also diminished in size when compared to the object.
(c) When light falls perpendicular to an interface of two media, it goes straight forward without bending. This could be explained by Fermat’s principle that states that the light will take that path that uses the least time.
Note:
Image formation is quite simple if you keep some important points in mind. Other than the two rules used in the solution, we also have another rule. This rule states that the ray going towards focus on another side becomes parallel to the principal axis after refraction.
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