A) 0.55 cm towards the lens
B) 0
C) 0.55 cm away from the lens
D) 1.1 cm away from the lens
Correct Answer: C
Solution :
\[U\,\,\,=\,\,-10\] \[V\,\,=\,\,10\] \[f\text{ }=\text{ }5\text{ }cm\] Shift due to slab = \[t\left( 1-\frac{1}{\mu } \right)\] \[=\,\,\,\,1.5\left( 1-\frac{1}{1.5} \right)\] \[=\,\,\,\,1.5\,\,\times \,\,\frac{1}{3}\,\,=\,\,\,0.5\,cm\] Now for lens \[u\,\,=\,\,-\,9.5\,cm\] \[f\,\,=\,\,5\,cm\] \[\frac{1}{v}\,\,-\,\,\frac{1}{u}\,\,=\,\,\frac{1}{f}\] \[\frac{1}{v}\,\,-\,\,\frac{1}{-9.5}\,\,=\,\,\frac{1}{5}\] \[\frac{1}{v}\,\,=\,\,\frac{1}{5}\,\,=\,\,\frac{1}{9.5}\] \[v\,\,=\,\,\frac{5\,\,\times \,\,9.5}{4.5}\] \[v=\text{ }10.55\text{ }cm\] So, screen is shifted away from lens by 0.55 cm.You need to login to perform this action.
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