91影视

Focal length is characteristic of the lens and gives the idea of how clearly and with what magnitude the image will be displayed by the lens. When an object is placed at infinity the image will be formed at the focal length.

Magnification of a lens is nothing but the lens鈥檚 capability to magnify the image of any regular size object. Mathematically,

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{y}\text{'}}{\mathrm{y}} \\ =-\frac{\mathrm{v}}{\mathrm{u}} \end{gathered}$$

As, the image obtained is erect so it is inferred that the image size is positive. Therefore, the magnification is also positive.

The expression in this case connecting the image distance v, the object distance u and the focal length f is,

$\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}$ ...(i)

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{y}\text{'}}{\mathrm{y}} \\ =\frac{1.30\mathrm{cm}}{0.400\mathrm{cm}} \\ =+3.25 \end{gathered}$$

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\ =+3.25 \\ \mathrm{v}=+3.25\mathrm{u} \end{gathered}$$

Putting the values in equation (i),

Solving the equation for u and v,

$$\begin{gathered} \mathrm{u}=6.23\mathrm{cm} \\ \mathrm{v}=-\left(6.23\mathrm{cm}\right)\left(3.25\right) \\ =-20.2\mathrm{cm} \end{gathered}$$

So, the virtual image is 20.2 cm left of the lens and the object is right to the 6.23 cm lens.