A tank is filled with benzene to a height of 120 mm . The apparent depth of a needle lying at a bottom of the tank is measured by a microscope to be 80 mm . Then the refractive index of benzene will be

Question

A tank is filled with benzene to a height of 120 mm . The apparent depth of a needle lying at a bottom of the tank is measured by a microscope to be 80 mm . Then the refractive index of benzene will be (A)  4.5  (B)  3  (C)  .75  (D)  1.5

Tardigrade Admin · Accepted Answer

Real depth d1=120 mm Apparent depth da=80 mm Refractive index of benzene is given by μ =(real depth/apparent depth)=(d1/da)=(120/80)=1.5

Q. A tank is filled with benzene to a height of $120 mm$ . The apparent depth of a needle lying at a bottom of the tank is measured by a microscope to be $80 mm$ . Then the refractive index of benzene will be

$4.5$

$3$

$.75$

$1.5$

Real depth $d_{1} = 120 mm$ Apparent depth $d_{a} = 80 mm$ Refractive index of benzene is given by $μ = \frac{re a l d e pt h}{a pp a re n t d e pt h} = \frac{d _{1}}{d _{a}} = \frac{120}{80} = 1.5$

Q. A tank is filled with benzene to a height of 120mm . The apparent depth of a needle lying at a bottom of the tank is measured by a microscope to be 80mm . Then the refractive index of benzene will be

4.5

3

.75

1.5

Real depth d1​=120mm Apparent depth da​=80mm Refractive index of benzene is given by μ=apparentdepthrealdepth​=da​d1​​=80120​=1.5

Q. A tank is filled with benzene to a height of $120 mm$ . The apparent depth of a needle lying at a bottom of the tank is measured by a microscope to be $80 mm$ . Then the refractive index of benzene will be

$4.5$

$3$

$.75$

$1.5$

Real depth $d_{1} = 120 mm$ Apparent depth $d_{a} = 80 mm$ Refractive index of benzene is given by $μ = \frac{re a l d e pt h}{a pp a re n t d e pt h} = \frac{d _{1}}{d _{a}} = \frac{120}{80} = 1.5$