Question

Convex lens focusses distant object 40 cm from it on screen placed 10 cm away from it. glass plate of RI =1.5 and thickness 3 cm is placed between lens and screen where should object be placed so that image is again focussed on screen

Answer 1

Dear student
$$\begin{gathered} U\sin glensformula \\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{10}-\frac{1}{-40}=\frac{1}{f} \\ \frac{1}{f}=\frac{5}{40}=\frac{1}{8} \\ f=8cm \\ shiftproducedbyglassslabinthedirrectionofincidentray=t\times (1-\frac{1}{\mu })=3\times (1-\frac{1}{1.5})=1cm \\ Fortheimagetoformonthescreenagainnowlensshouldformtheimagejust1cmbeforethescreensoafterrefractionfromtheslabthatby1cmfinalimageisagainformedonthescreen \\ imagedis\tan cefromfromlensis9cm \\ U\sin glensformula \\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{9}-\frac{1}{x}=\frac{1}{8} \\ \frac{1}{x}=\frac{1}{9}-\frac{1}{8} \\ \frac{1}{x}=\frac{-1}{72} \\ x=-72 \\ soobjectlies72cminfrontoflens \\ Regards \end{gathered}$$