The question asks about the relationship between the power of a lens and its focal length. Let's explore this concept.
In optics, the power of a lens is a measure of how strongly it converges or diverges light. It is defined as the reciprocal of the focal length of the lens.
The mathematical relationship between the power (\(P\)) of a lens and its focal length (\(f\)) is given by the formula:
$P = \frac{1}{f}
Where:
It's important that the focal length is in meters for the power to be in diopters. If the focal length is given in centimeters, it must be converted to meters before calculating the power.
Two quantities are directly proportional if they change by the same factor. If quantity A is directly proportional to quantity B, it means that as B increases, A increases proportionally, and as B decreases, A decreases proportionally. Mathematically, this is represented as \(A \propto B\).
Looking at the formula \(P = \frac{1}{f}\), we can see how the power (\(P\)) relates to the focal length (\(f\)).
Therefore, the power (\(P\)) is directly proportional to the term \(\frac{1}{f}\). This can be written as:
$P \propto \frac{1}{f}
Based on the definition and the formula \(P = \frac{1}{f}\), the power of a lens is directly proportional to the reciprocal of its focal length (\(\frac{1}{f}\)).
What special name is given to the frictional force exerted by a fluid?
______ is used in periscope.
Zero degree centigrade is equal to what degree Fahrenheit?
A. 100°F
B. 30°F
C. 34°F
D. 32°F
Keeping voltage constant, if more lamps are put into a series circuit, the overall current in the circuit:
A. Increases
B. Decreases
C. Remains the same
D. Becomes infinite
Excessive curvature of eye lens leads to _______