A) 25 cm left of \[{{S}_{4}},\]virtual
B) 25 cm right of \[{{S}_{4}},\]real
C) 15 cm left of \[{{S}_{4}},\]virtual
D) 20 cm right of \[{{S}_{4}},\]virtual
Correct Answer: A
Solution :
[a] We will have single surface refractions successively at the four surfaces \[{{S}_{1}},\text{ }{{S}_{2}},\text{ }{{S}_{3}}\,\]and \[{{S}_{4}}.\] Do not forget to shift origin to the vertex of respective surface. \[\frac{1.5}{{{v}_{1}}}-\frac{1}{\infty }=\frac{\left( 1.5-1 \right)}{+10}\text{ }{{v}_{1}}=30cm\] First image is object for the refraction at second surface. For refraction at surface \[{{S}_{2}}\]: Light travels from glass to air \[\frac{1}{{{v}_{2}}}-\frac{1.5}{\left( +25 \right)}=\frac{1-1.5}{\left( + \right)5}\text{ }{{\text{v}}_{2}}=-25cm\] For refraction at surface\[{{S}_{4}}\]: Light travels from air to glass. \[\frac{1.5}{{{v}_{3}}}-\frac{1}{\left( -35 \right)}=\frac{\left( 1-1.5 \right)}{\left( -5 \right)}\text{ }{{\text{v}}_{3}}=-35/3cm\] For refraction at surface\[{{S}_{4}}\]: Light travels from glass to air similarly, \[{{V}_{4}}=-25cm\] The final image is virtual, formal at 25 cm to the left of the vertex of surface\[{{S}_{4}}\].
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