Answer :
Final answer:
Using the Lensmaker's Equation, we can deduce that four lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and 8.00 cm, by forming four different surface configurations.
Explanation:
To answer your question on how many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and 8.00 cm in absolute magnitude, we will need to use the Lensmaker's Equation. The Lensmaker's Equation essentially relates the focal length of a lens to the radii of curvature of its surfaces and its refractive index.
In the case of simple lenses in air, the Lensmaker's Equation is given as follows: 1/f = (n - 1)(1/R1 - 1/R2), where 'f' is the focal length, 'n' is the refractive index, and 'R1' and 'R2' are the radii of curvature of the lens surfaces.
Assuming the refractive index of the material of the lens does not change and replacing 'R1' and 'R2' with 4.00 cm and 8.00 cm, note that if the two surfaces are arranged in different combinations - for instance, two convex surfaces, two concave surfaces, one convex and one concave, and vice-versa - you will get four different configurations. Thus, by manipulating these surfaces, you can obtain four different focal lengths.
Learn more about Lensmaker's Equation here:
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Answer:
The lenses with different focal length are four.
Explanation:
Given that,
Radius of curvature R₁= 4
Radius of curvature R₂ = 8
We know ,
Refractive index of glass = 1.6
When, R₁= 4, R₂ = 8
We need to calculate the focal length of the lens
Using formula of focal length
[tex]\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})[/tex]
Put the value into the formula
[tex]\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})[/tex]
[tex]\dfrac{1}{f}=\dfrac{9}{40}[/tex]
[tex]f=4.44\ cm[/tex]
When , R₁= -4, R₂ = 8
Put the value into the formula
[tex]\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})[/tex]
[tex]\dfrac{1}{f}=-\dfrac{3}{40}[/tex]
[tex]f=-13.33\ cm[/tex]
When , R₁= 4, R₂ = -8
Put the value into the formula
[tex]\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})[/tex]
[tex]\dfrac{1}{f}=\dfrac{3}{40}[/tex]
[tex]f=13.33\ cm[/tex]
When , R₁= -4, R₂ = -8
Put the value into the formula
[tex]\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})[/tex]
[tex]\dfrac{1}{f}=-\dfrac{9}{40}[/tex]
[tex]f=-4.44\ cm[/tex]
Hence, The lenses with different focal length are four.
