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Physics Question 50 – JEE-MAIN 2025

A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm. The value of H is _______ cm. (Consider liquids are immiscible)

When an object is viewed through a medium of different refractive index, its apparent position shifts.

Step 1: Identify Given Parameters and Formula✦ Active

We are given two immiscible liquid layers. For the first liquid, height h1=60 cm and refractive index n1=1.2. For the second liquid, height h2=H cm and refractive index n2=1.6. The total apparent shift in the position of the bottom of the container, when viewed from above, is Stotal=40 cm. The formula for apparent shift S due to a single layer of thickness h and refractive index n (when viewed from air) is S=h(11n). For multiple layers, the total shift is the sum of individual shifts.

Step 2: Calculate Individual Shifts○ Expand

The shift due to the first liquid layer (S1) is:

S1=h1(11n1)=60(111.2)=60(11012)=60(156)=60(16)=10 cm

The shift due to the second liquid layer (S2) is:

S2=h2(11n2)=H(111.6)=H(11016)=H(158)=H(38)
💡 Teacher's Secret Hint

Ensure correct calculation of the term (11/n) for each liquid.

Step 3: Sum Shifts and Solve for H○ Expand

The total apparent shift is the sum of the individual shifts:

Stotal=S1+S2

Substitute the given total shift and the calculated individual shifts:

40=10+H(38)

Now, solve for H:

30=H(38)H=30×83H=10×8H=80 cm
💡 Teacher's Secret Hint

Double-check the algebraic manipulation to isolate H.

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