When an object is placed between F and 2F of a convex lens the image formed is smaller and upright?

An object is placed between f and 2f of a convex lens. Which of the following statements correctly describes its image ? a real, larger than the object c inverted, same size as object

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Answer

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Hint: To understand what happens when an object is placed between F and 2F we have to draw the ray diagram. Placing the object at different positions will enable us to understand how the size of the image varies with the object. If the object is placed at the focus, the image will be formed at infinity and the object is placed at 2F the image will be formed at 2F on the other side of the lens.

Complete answer:

To begin with let us draw the ray diagram placing the same object but at two different positions at between F and 2F.

From the above ray diagram we can observe that If we place the object between F(focus)and 2F the image formed is always beyond 2F. The images are real and always inverted.Let us say an object is placed at 2F from the optical centre. Let us find its image using the lens formula for a convex lens i.e.$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ where f is the focal length of the lens, v is the image distance and u is the object distance Now let us place an object at u=-2f i.e. the centre of curvature. The minus sign indicates that the distance measured from the optical centre is opposite to the direction of the ray of light. Using the lens formula to let's calculate its image distance.$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{-2f}$ Multiplying 2 on both the sides of the above equation we get,$\dfrac{2}{f}=\dfrac{2}{v}+\dfrac{1}{f}$ $\dfrac{2}{f}-\dfrac{1}{f}=\dfrac{2}{v}$ taking f common in the denominator$\dfrac{1}{f}=\dfrac{2}{v}$ hence v=2f The magnification for a convex lens is given by $m=-\dfrac{v}{u}......(1)$ When the object is placed at the centre of curvature, image distance is equal to object distance and from equation 1 we can conclude that there is no magnification of the image. But if the image is placed between 2F and F its object distance keeps on reducing but its image distance keeps on increasing . Hence the image gets magnified.

So, the correct answer is “Option d”.

Note:

In ray optics the distance is always measured from the optical centre to any given point. The distance measured is taken as negative if the path traversed from the optical centre to a given point is in the opposite direction to that of the incident ray. The magnification formula for a convex lens has a minus sign which indicates that the object is inverted.