Question

A concave lens has a focal length of $20cm$ at what distance a $5cm$ tall object is placed so that it from an image at $15cm$ from the lens also calculates the size of the image formed.

Answer 1

The magnification equation has a relation with the ratio of the image distance and objects distance to the ratio of the image height and object height.
The mirror equation gives the quantitative relationship between the object distance, the image distance, and the focal length.
These two equations are used to solve this problem.

Complete step-by-step solution:
According to the given question,
$v = - 15\;cm$(which is negative because of the convex lens)

$$f = - 20\;cm$$ Let us consider the object distance as $$u$$ Now the lens formula is, $$\dfrac{1}{V} - \dfrac{1}{u} = \dfrac{1}{f}$$ Now substitute the value in the equations we get, $$\dfrac{1}{{ - 15}} - \dfrac{1}{u} = \dfrac{1}{{ - 20}}$$ Now rearranging the above equation, $$\dfrac{1}{u} = \dfrac{1}{{20}} + \dfrac{1}{{ - 15}}$$ Hence it becomes, $$\dfrac{1}{u} = \dfrac{{3 - 4}}{{60cm}} = \dfrac{1}{{ - 60}}$$ From this $$u$$ is, $$u = - 60\;cm$$ Here, the object is placed $$60\;cm$$away from the lens on the same side as the image. Now, $${h_1} = 5cm$$ The magnification is, $$m = \dfrac{{{h_2}}}{{{h_1}}} = \dfrac{v}{u}$$ Now apply the value of $$v$$ and $$u$$, $$ \Rightarrow \dfrac{{{h_2}}}{{5cm}} = \dfrac{1}{4}$$ $$ \Rightarrow {h_2} = \dfrac{{5cm}}{4}$$ Now the value of $${h_2}$$ is $${h_2} = 1.25cm$$ If the sign is positive the image is virtual and erect. Finally, the image formed at $${h_2} = 1.25cm$$ **Note:** Ray diagrams can be used to determine the image location, size, orientation, and type of image formed of From the Ray diagrams, many useful pieces of information about object-image relationships are gained. It fails to provide the information in a quantitative form. While a ray diagram may help one determine the approximate location and size of the image, it will not provide numerical information about image distance and object size.