a particular camera has a focal length of 50 millimeters (mm). a photographer wants to get a closeup of…

a particular camera has a focal length of 50 millimeters (mm). a photographer wants to get a closeup of geological formation just under a foot away or at a value of 0.275 meter. a standard equation applies as follows to find image distance (i) involving object distance (o) and focal length (f).\n\\(\\frac{1}{i}=\\frac{1}{f}-\\frac{1}{o}\\)\nsince the focal lengths are expressed in mm, all values will be included in the formula in mm.\n\\(\\frac{1}{i}=\\frac{1}{50}-\\frac{1}{275}\\)\nwhat is the lowest value that can be used for the denominator to complete the equation?

a particular camera has a focal length of 50 millimeters (mm). a photographer wants to get a closeup of geological formation just under a foot away or at a value of 0.275 meter. a standard equation applies as follows to find image distance (i) involving object distance (o) and focal length (f).\n\\(\\frac{1}{i}=\\frac{1}{f}-\\frac{1}{o}\\)\nsince the focal lengths are expressed in mm, all values will be included in the formula in mm.\n\\(\\frac{1}{i}=\\frac{1}{50}-\\frac{1}{275}\\)\nwhat is the lowest value that can be used for the denominator to complete the equation?

Answer

Explanation:

Step1: Find common denominator

To subtract $\frac{1}{50}-\frac{1}{275}$, we need a common denominator. We find the least - common multiple (LCM) of 50 and 275.

Step2: Prime - factorize the numbers

Prime - factorize 50 = $2\times5^{2}$ and 275 = $5^{2}\times11$.

Step3: Calculate the LCM

The LCM is $2\times5^{2}\times11= 550$.

Answer:

550