JEE Main & Advanced Physics Ray Optics Real and Apparent Depth

If object and observer are situated in different medium then due to refraction, object appears to be displaced from it's real position.

(1) When object is in denser medium and observer is in rarer medium

(i) \[\mu =\frac{\text{Real depth}}{\text{Apparent depth}}=\frac{h}{{{h}'}}\]

(ii) Real depth > Apparent depth

(iii) Shift \[d=h-{{h}^{'}}=\left( 1-\frac{1}{\mu } \right)\,h\].

For water \[\mu =\frac{4}{3}\Rightarrow d=\frac{h}{4}\];  For glass \[\mu =\frac{3}{2}\Rightarrow d=\frac{h}{3}\]

(iv) Lateral magnification : consider an object of height \[x\] placed vertically in a medium \[{{\mu }_{1}}\] such that the lower end \[(B)\] is a distance h from the interface and the upper end (A) at a distance \[(h-x)\] from the interface.

Distance of image of B (i.e. B') from the interface = \[\frac{{{\mu }_{2}}}{{{\mu }_{1}}}h\]

Distance of image of A (i.e. A') from the interface \[=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}(h-x)\]

Therefore, length of the image \[\frac{{{\mu }_{2}}}{{{\mu }_{1}}}x\]

or, the lateral magnification of the object \[m=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\]\[=\frac{1}{\mu }\]

(v) If a beaker contains various immiscible liquids as shown then

Apparent depth of bottom\[=\frac{{{d}_{1}}}{{{\mu }_{1}}}+\frac{{{d}_{2}}}{{{\mu }_{2}}}+\frac{{{d}_{3}}}{{{\mu }_{3}}}+....\]

\[\mu \]combination = \[\frac{{{d}_{AC}}}{{{d}_{App.}}}=\frac{{{d}_{1}}+{{d}_{2}}+.....}{\frac{{{d}_{1}}}{{{\mu }_{1}}}+\frac{{{d}_{2}}}{{{\mu }_{2}}}+....}\]

(In case of two liquids if \[{{d}_{1}}={{d}_{2}}\] than \[\mu =\frac{2{{\mu }_{1}}{{\mu }_{2}}}{{{\mu }_{1}}+{{\mu }_{2}}}\])

(2) Object is in rarer medium and observer is in denser medium

(i)  \[\mu =\frac{{{h}^{'}}}{h}\]

(ii) Real depth < Apparent depth.

(iii) \[d=(\mu -1)h\]

(iv) Shift for water \[{{d}_{w}}=\frac{h}{3}\];  Shift for glass \[{{d}_{g}}=\frac{h}{2}\]