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We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What must be the focal length of the appropriate positive lens? If it is a bi-convex lens of refractive index 1.5, what are the values of radii of curvature?

Respuesta :

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

[tex]\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}[/tex]

[tex]\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}[/tex]

f = 30 cm

using lens formula

[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})[/tex]

[tex]R_1 = R\ and\ R_2 = -R[/tex]

[tex]\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})[/tex]

[tex]R = (n -1)\ f [/tex]

[tex]R = 2(1.5 -1)\ 30[/tex]

R = 30 cm

hence, the radii of curvature is 30 cm.

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