Question

If the focal length of a lens of a camera is 5 f and that of another is 2.5 f, what is the time of exposure for the second if for the first one $\frac{1}{200}s$ ? (where, f is focal length/unit aperture)

• A. $\frac{1}{200}$
• B. $\frac{1}{800}$
• C. $\frac{1}{6400}$
• D. $\frac{1}{3200}$

Answer 1

Answer: (B)

$\frac{1}{800}$

f-number for given first lens is $5$ and for second one is $2.5$

As we know

$\Rightarrow Aperture\left(A\right)\propto {\left(\frac{1}{f-number}\right)}^2$

$\Rightarrow Timeofexposure\left(t\right)=C{\left(\frac{1}{Aperture}\right)}^2...\left(1\right)$ where $C$ is a constant

$\Rightarrow Timeofexposure\left(t\right)\propto {\left(\frac{1}{Aperture}\right)}^2...\left(2\right)$

Using the equation together we get

$\Rightarrow Timeofexposure\left(t\right)\propto {\left(Aperture\right(A\left)\right)}^2$

$\Rightarrow Timeofexposure\left(t\right)=K{\left(Aperture\right(A\left)\right)}^2$ where $K$ is a constant

$\Rightarrow \frac{t_1}{t_2}=\frac{A_2^2}{A_1^2}$

$\Rightarrow t_2=t_1\frac{A_1^2}{A_2^2}$

Putting $t_1=\frac{1}{200}s$, $A_1=C{\left(\frac{1}{5}\right)}^2$,$A_2=C{\left(\frac{1}{2.5}\right)}^2$

$\Rightarrow t_2=\frac{1}{200}{\left(\frac{2.5}{5}\right)}^2\Rightarrow t_2=\frac{1}{800}s$

so the correct option is $\left(B\right)$.