Respuesta :
Answer:
[tex]f =17.15\ cm[/tex]
Explanation:
given,
refractive index of lens, n = 1.70
Radius of curvature of front surface. R₁ = 20 cm
Radius of curvature of the back surface, R₂ = 30 cm
focal length= ?
[tex]\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})[/tex]
R₁ = +20 cm
R₂ = -30 cm
n = 1.70
[tex]\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})[/tex]
[tex]\dfrac{1}{f}=0.70 \times 0.0833[/tex]
[tex]f = \dfrac{1}{0.7 \times 0.0833}[/tex]
[tex]f =17.15\ cm[/tex]
the focal length of the lens is equal to 17.15 cm
The focal length of the lens is f=17.15 cm.
Given:
Refractive index of lens, n = 1.70
Radius of curvature of front surface. R₁ = 20 cm
Radius of curvature of the back surface, R₂ = 30 cm
To find:
focal length= ?
[tex]\frac{1}{f}= (n-1)(\frac{1}{R_1} -\frac{1}{R_2})[/tex]
R₁ = +20 cm
R₂ = -30 cm
n = 1.70
On substituting these values, we will get:
[tex]\frac{1}{f} =(1.70-1)(\frac{1}{20} -\frac{1}{-30} )\\\\\frac{1}{f} =0.70*0.0833\\\\f=17.15 cm[/tex]
Therefore, the focal length of the lens is equal to 17.15 cm.
Learn more:
brainly.com/question/23340569
