Question

An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of the surface). The surface’s radius of curvature is 11 cm. Use these indices of refraction: nair = 1.0 and nglass = 1.5. show answer Incorrect Answer 50% Part (a) Find the refraction image distance, in centimeters. Include its sign.

Answer 1

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. $d_{i}=-0.0566cm$

b. $M=441.69$

Explanation:

For this type of numerical we will use the following formulas

$\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1} }{R}$......... Eq1

where,

$n_{1}$$=$refractive index of the medium surrounding refracting surface/object

i.e. $n_{1}$=$n_{air}$  

$n_{2}$$=$ refractive index of the refracting surface/object

i.e. $n_{2}=n_{glass}$

$d_{0}=$ distance of object from the vertex of the refracting surface

$d_{i}=$distance of image from the vertex of the refracting surface

$R=$radius of curvature of the refracting surface

$M=\frac{d_{0} }{d_{i} }$ ........... Eq2

where,

$M=$magnification

Convention:

$R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R$

Given:

$n_{1}$=$n_{air}$$=1.0$

$n_{2}=n_{glass}$$=1.5$

$d_{0}=$$25cm$

$R=-11$$cm$ because refraction surface is concave

Required:

a. $d_{i}=$$?$

b. $M=?$

Solution:

a. putting values in eq1, we get

$\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}$

$\frac{1.5}{d_{i} }=-0.045-0.040$

$d_{i}=(-0.085)(\frac{1}{1.5} )$

$d_{i}=-0.0566$cm

b. $M=\frac{25}{-0.05665}$

$M=441.69$