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Question: A convex lens of focal length f produces a real image x times the size of an object; then the distan...

A convex lens of focal length f produces a real image x times the size of an object; then the distance of the object from the lens is:

Explanation

Solution

By using the magnification formula, we can get the expression connecting the distance of the object from the mirror with the distance of the image from the mirror. Then by using the lens formula, we shall be able to obtain the required expression for distance of the object from the lens.
Formula used:
Lens formula for focal length and the distances of object and image from the lens is given as:
1f=1v1u\dfrac{1}{f} = \dfrac{1}{{\text{v}}} - \dfrac{1}{u}
where f is used to represent the focal length of the lens, while u and v represent the distance of the object and its image respectively from the lens.
Magnification is given in terms of the size of the object and its image by the following formula:
m=hiho=vum = \dfrac{{{h_i}}}{{{h_o}}} = \dfrac{{\text{v}}}{{ - u}}

Complete answer:
We are given a convex lens whose focal length is given as f. This lens produces a real image x times the size of an object. Therefore, using the magnification formula, we can write
m=hiho=vum = \dfrac{{{h_i}}}{{{h_o}}} = \dfrac{{\text{v}}}{{ - u}}
Here hi{h_i} is the height of the image of object while ho{h_o} is the height of the object; u denotes the distance of the object from the lens while v denotes the distance of the image from the lens.
We are given that
hi=x.h0 hiho=x \begin{gathered} {h_i} = x.{h_0} \\\ \Rightarrow \dfrac{{{h_i}}}{{{h_o}}} = x \\\ \end{gathered}
Inserting the given information in the magnification formula, we get
vu=x v=ux \begin{gathered} \dfrac{{\text{v}}}{{ - u}} = x \\\ \Rightarrow v = - ux \\\ \end{gathered}
Inserting this expression in the lens formula, we get
1f=1v1u 1f=1ux1u=1u(1x+1) u=f(x+1x) \begin{gathered} \dfrac{1}{f} = \dfrac{1}{{\text{v}}} - \dfrac{1}{u} \\\ \dfrac{1}{f} = \dfrac{1}{{ - ux}} - \dfrac{1}{u} = - \dfrac{1}{u}\left( {\dfrac{1}{x} + 1} \right) \\\ \Rightarrow u = - f\left( {\dfrac{{x + 1}}{x}} \right) \\\ \end{gathered}

Note:
1. All distances to the left of the lens are taken to be negative while all distances to the right of the lens are taken to be negative.
2. The focus is defined on the basis of curvature of the front circle of the lens. A concave lens has negative focus because it lies on the left side of the lens while a convex lens has positive focus because it lies on the right side of the lens.