The bottom of a beaker containing a liquid appears to rise by 4 cm . On increasing the depth of the liquid bu 12 cm . The bottom appears to rise by 7 cm. The refractive index of the liquid.
Answers
Answer:
The bottom of a beaker containing a liquid appears to rise by 4 cm. On increasing the depth of a liquid by 12 cm, the bottom appears to rise by 7 cm.
To find:
Refractive Index of liquid.
Calculation:
\sf{rise \: in \: image = real \: depth - apparent \: depth}riseinimage=realdepth−apparentdepth
\sf{ = > r= d - \dfrac{d}{ \mu} }=>r=d−
μ
d
\sf{ = > r= d (1- \dfrac{1}{ \mu} )}=>r=d(1−
μ
1
)
In 1st case:
\sf{ 4= d (1- \dfrac{1}{ \mu} ) \: \: \: \: \: ........(1)}4=d(1−
μ
1
)........(1)
In 2nd case:
\sf{ 7= (d + 12)(1- \dfrac{1}{ \mu} ) \: \: \: \: \: ........(2)}7=(d+12)(1−
μ
1
)........(2)
Dividing the two Equations:
\rm{ \therefore \dfrac{d + 12}{d} = \dfrac{7}{4} }∴
d
d+12
=
4
7
\rm{ = > 4d + 48 = 7d}=>4d+48=7d
\rm{ = >3d = 48 }=>3d=48
\rm{ = >d = 16 \: m }=>d=16m
Now, putting value of d in eq.(1):
\sf{ \therefore 4= 16 \times (1- \dfrac{1}{ \mu} ) }∴4=16×(1−
μ
1
)
\sf{ = > \dfrac{1}{4} = (1- \dfrac{1}{ \mu} ) }=>
4
1
=(1−
μ
1
)
\sf{ = > \dfrac{1}{ \mu} = 1 - \dfrac{1}{4} }=>
μ
1
=1−
4
1
\sf{ = > \dfrac{1}{ \mu} = \dfrac{3}{4} }=>
μ
1
=
4
3
\sf{ = > \mu= \dfrac{4}{3} }=>μ=
3
4
So, final answer is:
\boxed{ \mathfrak{ \red{ \mu= \dfrac{4}{3} }}}
μ=
3
4