91影视

The focal length is the distance between the convex or concave mirror and the focal point of the mirror.

The relation between the distance of object s, the distance of the image s' and the focal length is

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s^{\mathrm{鈥�}}}$

The ratio of the distance of the image s' to the ratio of the object s is known as magnification .

$m=-\frac{s\text{'}}{s^{\mathrm{鈥�}}}$

The ratio of real image y' and focal length$f_2$ is the distance of the image u' .

$s^{\mathrm{鈥�}}=\frac{y^{\mathrm{鈥�}}}{f_2}$

The ratio of real image y' and focal length$f_1$ is the distance of the object u.

$s^{\mathrm{鈥�}}=\frac{y^{\mathrm{鈥�}}}{f_1}$

From the graph the distance of the image is and $\frac{1}{m}$is lateral magnification.

So, by the magnification formula

s' = - ms

By the relation between the distance of object , the distance of the image and the focal length f , the distance of image s is:

$\begin{array}{r}\frac{1}{s}+\frac{1}{s^{\mathrm{鈥�}}}=\frac{1}{f}\\ \frac{1}{s}+\left(鈭�\frac{1}{ms}\right)=\frac{1}{f}\\ s=鈭�f\frac{1}{m}+f\end{array}$

So, the intercept the equation is and the slope is -f and graph given is linear.

From the equation $s=鈭�f\frac{1}{m}+f$

$\begin{array}{r}f=鈭�\text{slope}\\ =鈭�\left(\frac{50\mathrm{cm}鈭�38\mathrm{cm}}{鈭�1.0鈭�(鈭�0.5)}\right)\\ =24\mathrm{cm}\end{array}$

Hence, as the equation of curve$s=鈭�f\frac{1}{m}+f$ is linear, therefore the data plotted should be close to a straight line and the focal length of the mirror is 24cm.

Given that, y' = 8 mm

The lateral magnification is:

$\begin{array}{r}\frac{1}{m}=鈭�\frac{y}{y^{\mathrm{鈥�}}}\\ =鈭�\frac{4}{8}\\ =鈭�0.5\end{array}$

Now, the distance of object is:

$\begin{array}{r}s=鈭�f\frac{1}{m}+f\\ =鈭�\left(24\mathrm{cm}\right)(鈭�0.5)+24\mathrm{cm}\\ =36\mathrm{cm}\end{array}$

Since the distance is inversely proportional to the magnification. Therefore, with increase in magnification, there is increase in height of the image the object will be closer to the lens.

$s鈭�\frac{1}{m}$

Hence, if mirror鈥檚 vertex is placed 36 cm from the object, the image will be real 8.00 mm , tall, and inverted and the object will be closer to the mirror to increase the height of the inverted, real image.

As the distance of object approaches 25 cm the magnification increases until it reaches$鈭�$ . Thus 1/m will approach to zero. The magnification will be because is very close to the focal length.

Since the virtual image is formed on carboard when s < 25 cm . Therefore, image could not be sharp.

Hence, when the distance is approximately equal to the focal length therefore, 1 / m approaches zero as approaches 25 cm and since the virtual image is formed, therefore, the image cannot be sharp.