Question

Exercise 1 Assume that a container is filled with 4cm depth of water and an object is located at the bottom of the container The refraction index of water is n 1.33 Ask what will be the depth of the image of the object when we look at it on the top of the container? Solution: Based on Lens Makers equation, we obtain 1.33 1.0 1.0-1.33 d2- -3(cm) 4cm Water . Image is located 3cm under water. It is closer than the real object to our eyes.

Can someone explain this a little bit??

Answer 1

EXPLANATION

I hope you are aware of lens maker's formula ==>

$\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R}$

$u \rightarrow Distance\ of\ object\ from\ the\ lens\ surface = 4\ cm$

$v \rightarrow Distance\ of\ image\ from\ the\ lens\ surface$

$R\rightarrow Radius\ of\ the\ plane\ lens\ surface = infinity$

Therefore,

$\frac{4/3}{4} + \frac{1}{v} = \frac{1 - 4/3}{infinity} = 0$ (n₂ = 1 because the medium outside is air and refractive index of air is 1)

$\Rightarrow {v} = -3\ cm$

This means that the image is formed inside the water rather than air. And the depth of image is 3 cm. ==> ANSWER