A student placed a 4 cm tall object in front of a concave mirror to obtain an image at a distance of 30 cm from the mirror. Find the distance at which the student has placed the object. Also, find the height of the image obtained.

                                                         (OR)

A real image $\frac{3^{\mathrm{th}}}{4}$ of the size of an object is formed by a convex lens when the object is at a distance of 15 cm from it. Find the focal length of the lens.

According to the question, 

Image distance, 𝑣 =− 30 𝑐𝑚

Focal length 𝑓 =− 20 𝑐𝑚 

Height ℎ = 4 𝑐𝑚 

Let us consider object distance 𝑢 =−? 

Size of the image formed ℎ' =−? 

According to the mirror formula

$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$

Put value of 𝑓, 𝑣 𝑎𝑛𝑑 𝑢 in above formula

$$\begin{gathered} \Rightarrow -\frac{1}{20}=-\frac{1}{30}+\frac{1}{\mathrm{u}} \\ \Rightarrow \frac{1}{\mathrm{u}}=-\frac{1}{20}+\frac{1}{30} \\ \Rightarrow \frac{1}{\mathrm{u}}=\frac{1}{30}-\frac{1}{20} \\ \Rightarrow \frac{1}{\mathrm{u}}=\frac{2-31}{60} \\ \Rightarrow \frac{1}{\mathrm{u}}=-\frac{1}{60} \\ \Rightarrow \mathrm{u}=-60 \mathrm{cm} \end{gathered}$$

Therefore, an object is placed at 60 𝑐𝑚 away from the mirror, on the same side of image. According to linear magnification,

$$\begin{gathered} \frac{-\mathrm{v}}{\mathrm{u}}=\frac{{\mathrm{h}}^{\text{'}}}{\mathrm{h}} \\ \Rightarrow \frac{-(-30)}{-60}=\frac{{\mathrm{h}}^{\text{'}}}{4} \\ \Rightarrow \frac{30}{-60}=\frac{{\mathrm{h}}^{\text{'}}}{4} \\ \Rightarrow -\frac{1}{2}=\frac{{\mathrm{h}}^{\text{'}}}{4} \\ \Rightarrow -\frac{4}{2}={\mathrm{h}}^{\text{'}} \\ \Rightarrow {\mathrm{h}}^{\text{'}}=-2 \mathrm{cm} \end{gathered}$$

Thus, the height of the image is 2 𝑐𝑚.

                                          (OR)

Given, for the real and inverted images, the magnification will be negative.

$\text{ So, }\mathrm{m}=\frac{-3}{4}\text{ and }\mathrm{u}=-15 \mathrm{cm}$

As we know,

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\ \Rightarrow \frac{-3}{4}=\frac{\mathrm{v}}{-15} \\ \Rightarrow \mathrm{v}=\frac{-3\times -15}{4} \\ \Rightarrow \mathrm{v}=\frac{45}{4} \end{gathered}$$

Now,

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ \Rightarrow \frac{1}{\mathrm{f}}=\frac{4}{4/45}-\frac{1}{-15} \\ \Rightarrow \frac{1}{\mathrm{f}}=\frac{4}{45}+\frac{1}{15} \\ \Rightarrow \frac{1}{\mathrm{f}}=\frac{4+3}{45} \\ \Rightarrow \frac{1}{\mathrm{f}}=\frac{7}{45} \\ \Rightarrow \mathrm{f}=\frac{45}{7} \\ \Rightarrow \mathrm{f}=6.43 \mathrm{cm} \end{gathered}$$

Hence, the focal length of the lens is 6. 43 𝑐𝑚.