Question

A small piece of paper is stuck on a glass of radius r cm and is observed through the glass from
a position directly opposite. If a μ g = 3/2, then from the front surface of the glass, the image is​

Answer 1

Given:

The refractive index of glass n=1.5

The refractive index of air n'=1

To find:

The image magnification and position of it.

Solution:

Radius of curvature R

u= -2R

We know the formula of mirror

n'/v -n/u=n'-n/R

1/v -1.5/u=1-1.5/R

1/v-1.5/(-R)=-0.5/R

1/v=0.5/R - 1.5/R

1/v=-1/4R

v=-4R

Now the magnification

m=-v/u

m=4R/2R

m=2

Therefore the magnification and positioning of image is at 2

Answer 2

Explanation:

Given:

The refractive index of glass n=1.5 The refractive index of air n'=1

Let me find it:-

The image magnification and position of it.

Solution:

Radius of curvature R

u= -2R

We will use the formula of

$n'/v -n/u=n'-n/R</p><p>$

1/v -1.5/u=1-1.5/R

1/v-1.5/(-R)=-0.5/R

1/v=0.5/R - 1.5/R

1/v=-1/4R

v=-4R

Next step:-

Now the magnification

m=-v/um=2

Required answer:- ✅

• Therefore the magnification and positioning of image is at 2

• Hope it helps you mate:-
• #Be brainly

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