Question

What is real depth of a swimming pool when it's bottom appears to be raised by 1 m. Given refractive index of water is 4/3

Answer 1

Hey!

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Real depth of swimming pool =?
Refractive index of water = 4/3

Here,

n = real depth(x) / apparant depth(y)

x = ? , (x-y) = 1 m
n = 4/3

n = x/y

n-1 = x/y -1

n-1 = x-y/y

4/3 - 1 = 1/y

y = 3

Now, From,

x-y = 1
x= 1 + y
x = 1+3
x = 4m

Thus, real depth of swimming pool = 4 m

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Hope it helps...!!!

Answer 2

HELLO DEAR,

let the real depth of swimming pool = x
AND, apparent depth = y


Given that:-

Refractive index of water = 4/3
(X - y) = 1 ---------(1)
n = 4/3

Now,

We know that:-


n = real depth/ apparant depth = x/y


⇒n = x/y

⇒n-1 = x/y -1
∴ [ subtracting both side by (1) ]


we get,

⇒n-1 = x-y/y
∴ [ x - y = 1(given) ]

⇒4/3 - 1 = 1/y
∴ [ n = 4/3 ]


⇒y = 3 put in ---(1)


we get,

X - y = 1

⇒x - 1 = 3

⇒x = 3 + 1

⇒x = 4m

HENCE ,the real depth of swimming pool = 4 m


I HOPE ITS HELP YOU DEAR,
THANKS