How can we change a camera from F/4 to F/5.6 ?
A) Increase the aperture to 2 times keeping the focal distance constant.
B) Increase the aperture to \[\sqrt 2 \] time keeping the focal distance constant.
C) Increase the aperture to \[\dfrac{1}{2}\] time keeping the focal distance constant.
D) Increase the aperture to $\dfrac{1}{{\sqrt 2 }}$ time keeping the focal distance constant.
Answer
345.3k+ views
Hint: For solving this question, we have to use the concept of f-number or focal ratio which is the ratio of the system’s focal distance to the diameter of aperture.
${\text{N = }}\dfrac{f}{D}$
Where, N is the f- number, D is the diameter of the aperture and f is the focal distance.
Complete step by step answer:
Let us first understand this concept of f-number. The formula is given by :-
${\text{N = }}\dfrac{f}{D}$ ……. (1)
where N is the f-number, f is the focal distance and D is the diameter of aperture of the camera. The f-number (N) is represented by F/N.
Now, in the question given above we have to convert a camera from F/4 to F/5.6. Hence, it’s f-number changes from 4 to 5.6.
For F/4 camera, let’s ${{\text{D}}_1}$ be its diameter of aperture and f-number is 4, so using the equation (1),
$4\, = \,\dfrac{f}{{{{\text{D}}_1}}}$ …….(2)
For F/5.6 camera, let’s \[{{\text{D}}_2}\] be it’s diameter of aperture and f-number is 5.6, so using the equation (1),
$5.6\, = \,\dfrac{f}{{{{\text{D}}_2}}}$ …….(3)
Since the focal distance is constant in both the cases, we will equate the equations (2) and (3) as,
$\implies 4 \times {{\text{D}}_1}\, = \,\,5.6 \times {{\text{D}}_2} \\
\implies \dfrac{{{{\text{D}}_1}}}{{{{\text{D}}_2}}}\, = \,\,\dfrac{{5.6}}{4} \\
\implies \dfrac{{{{\text{D}}_1}}}{{{{\text{D}}_2}}}\, = \,\,\sqrt 2 \\
\therefore {{\text{D}}_2}\, = \,\,\dfrac{{{{\text{D}}_1}}}{{\sqrt 2 }} \\
$
Hence, the diameter of aperture of the F/5.6 camera should be $\dfrac{1}{{\sqrt 2 }}$ time the diameter of the aperture of the F/4 camera keeping the focal distance constant.
Therefore, (D) option is correct.
Note: The f-number explains the light gathering ability of lens only for the objects at an infinite distance away. In some cases, a working f-number is used. But make sure to use the correct formula because many students assume the formula to be $f = \dfrac{N}{D}$ instead of ${\text{N = }}\dfrac{f}{D}$
${\text{N = }}\dfrac{f}{D}$
Where, N is the f- number, D is the diameter of the aperture and f is the focal distance.
Complete step by step answer:
Let us first understand this concept of f-number. The formula is given by :-
${\text{N = }}\dfrac{f}{D}$ ……. (1)
where N is the f-number, f is the focal distance and D is the diameter of aperture of the camera. The f-number (N) is represented by F/N.
Now, in the question given above we have to convert a camera from F/4 to F/5.6. Hence, it’s f-number changes from 4 to 5.6.
For F/4 camera, let’s ${{\text{D}}_1}$ be its diameter of aperture and f-number is 4, so using the equation (1),
$4\, = \,\dfrac{f}{{{{\text{D}}_1}}}$ …….(2)
For F/5.6 camera, let’s \[{{\text{D}}_2}\] be it’s diameter of aperture and f-number is 5.6, so using the equation (1),
$5.6\, = \,\dfrac{f}{{{{\text{D}}_2}}}$ …….(3)
Since the focal distance is constant in both the cases, we will equate the equations (2) and (3) as,
$\implies 4 \times {{\text{D}}_1}\, = \,\,5.6 \times {{\text{D}}_2} \\
\implies \dfrac{{{{\text{D}}_1}}}{{{{\text{D}}_2}}}\, = \,\,\dfrac{{5.6}}{4} \\
\implies \dfrac{{{{\text{D}}_1}}}{{{{\text{D}}_2}}}\, = \,\,\sqrt 2 \\
\therefore {{\text{D}}_2}\, = \,\,\dfrac{{{{\text{D}}_1}}}{{\sqrt 2 }} \\
$
Hence, the diameter of aperture of the F/5.6 camera should be $\dfrac{1}{{\sqrt 2 }}$ time the diameter of the aperture of the F/4 camera keeping the focal distance constant.
Therefore, (D) option is correct.
Note: The f-number explains the light gathering ability of lens only for the objects at an infinite distance away. In some cases, a working f-number is used. But make sure to use the correct formula because many students assume the formula to be $f = \dfrac{N}{D}$ instead of ${\text{N = }}\dfrac{f}{D}$
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