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.

 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

d_o is the distance of the object  

d_i is the distance of the image

 d₀ = -31 cm

now, magnification is calculated using formula

   m = 1.254

Hence, magnification of the object is equal to 1.254

your answer would be fusion