Question

The rectangular metal tank is filled with a liquid of refractive index $\mu$. The observer, whose eye is in level with the top of the tank, can just see corner B. The value of $\mu$ is:

• A. $\frac{7}{4}$
• B. $\frac{5}{3}$
• C. $\frac{3}{2}$
• D. $\frac{5}{4}$
• E. 1/2
• F. 1

Answer 1

Answer: (D)

5/4

Answer 2

The problem describes a rectangular tank filled with a liquid of refractive index $\mu$. An observer, whose eye is at the level of the top of the tank, can just see corner B. We need to find the value of $\mu$.

1. Identify the Geometry and Ray Path:

• Let the dimensions of the tank be:
  • Height (depth of liquid) = 3m (vertical distance AB or CD).
  • Width = 4m (horizontal distance BC or AD).
• The observer's eye is at point D, which is the top-left corner of the tank.
• Corner B is the bottom-right corner of the tank.
• The phrase "can just see corner B" implies that the light ray from B, after refracting at the liquid-air interface, grazes the surface as it enters the observer's eye at D. This means the angle of refraction in air is $90^\circ$.
• Since the observer's eye is exactly at point D (a point on the liquid surface), the light ray from corner B must travel straight to point D within the liquid, and then emerge into the air at D.

2. Set up Coordinates: Let's place the bottom-left corner C at the origin (0,0).

• C = (0,0)
• B = (4,0) (since width BC = 4m)
• D = (0,3) (since height CD = 3m)
• A = (4,3)

3. Determine the Angle of Incidence ($i$):

• The light ray travels from B(4,0) to D(0,3) within the liquid.
• At point D, the liquid-air interface is horizontal (y=3). The normal to this surface is a vertical line.
• The angle of incidence, $i$, is the angle between the incident ray BD and the normal at D.
• Consider the right-angled triangle BCD.
  • BC = 4m (horizontal distance)
  • CD = 3m (vertical distance)
  • The hypotenuse BD is the path of the light ray in the liquid.
• The normal at D points vertically downwards (or upwards). The angle $i$ is the angle between the ray BD and the vertical line CD.
• In $\triangle BCD$, $\tan(\angle BDC) = \frac{\text{Opposite side}}{\text{Adjacent side}} = \frac{BC}{CD} = \frac{4}{3}$.
• Therefore, the angle of incidence $i = \angle BDC$.
• To find $\sin i$, we use the Pythagorean theorem: Hypotenuse $BD = \sqrt{BC^2 + CD^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$m.
• $\sin i = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{BD} = \frac{4}{5}$.

4. Determine the Angle of Refraction ($r$):

• As established, since the observer "can just see corner B" from point D (which is on the surface), the refracted ray in air must be grazing the surface.
• This means the angle of refraction $r = 90^\circ$.
• So, $\sin r = \sin 90^\circ = 1$.

5. Apply Snell's Law: Snell's Law states: $\mu_1 \sin i = \mu_2 \sin r$ Here, $\mu_1 = \mu$ (refractive index of liquid) and $\mu_2 = 1$ (refractive index of air).

• $\mu \sin i = 1 \cdot \sin r$
• Substitute the values: $\mu \left(\frac{4}{5}\right) = 1 \cdot (1)$
• $\mu = \frac{5}{4}$

The value of the refractive index $\mu$ is $\frac{5}{4}$.