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An object is located in air, 31 cm from the vertex of the convex surface of a block of glass (as viewed from the air side of the surface). The surface’s radius of curvature is 79 cm. Use these indices of refraction: nair = 1.0 and nglass = 1.5. (The Expert TA)

What is the magnification of the image? Include the magnification’s sign to indicate the orientation of the image with respect to the object.

Respuesta :

Answer:

 m = 1.254

Explanation:

given,

object distance = 31 cm

radius of curvature,R = 79 cm

refractive index of air, n₁ = 1

refractive index of glass,n₂ = 1.5

magnification of the image = ?

Relation of the image formed from refraction

[tex]\dfrac{n_1}{d_0}+\dfrac{n_2}{d_i}=\dfrac{n_2-n_1}{R}[/tex]

d_o is the distance of the object  

d_i is the distance of the image

[tex]d_i = \dfrac{n_2}{\dfrac{n_2-n_1}{R}}-\dfrac{n_1}{d_0}[/tex]

 d₀ = -31 cm

[tex]d_i = \dfrac{1.5}{\dfrac{1.5-1}{79}-\dfrac{1}{(-31)}}[/tex]

[tex]d_i = 38.87\ cm[/tex]

now, magnification is calculated using formula

[tex]m=\dfrac{-d_i}{d_o}[/tex]

[tex]m=\dfrac{-38.87}{-31}[/tex]

   m = 1.254

Hence, magnification of the object is equal to 1.254

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