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An object is placed at \mathrm{21\ cm} in front of a concave mirror of radius of a curvature \mathrm{10\ cm}. A glass slab of thickness \mathrm{3\ cm} and \mathrm{\mu =1.5} is the then placed close to the mirror in the space between the object and the mirror. The position of final image formed is:

Option: 1

\mathrm{-3.94\ cm}


Option: 2

\mathrm{4.3\ cm}


Option: 3

\mathrm{-4.93\ cm}


Option: 4

\mathrm{3.94\ cm}


Answers (1)

best_answer

Due to glass slab, increase in path
\mathrm{\begin{gathered} =(\mu-1) \mathrm{t}=(1.5-1) 3 \mathrm{~cm}=1.5 \mathrm{~cm} \\ \therefore \quad \mathrm{u}^{\prime}=-(21+1.5) \mathrm{cm} \quad \text { or } \quad \mathrm{u}^{\prime}=-22.5 \mathrm{~cm} \end{gathered}}
\mathrm{\text { Now, } \frac{1}{-22.5}+\frac{1}{v}=\frac{1}{-5}}
\text { or } \quad \frac{1}{\mathrm{v}}=\frac{1}{22.5}-\frac{1}{5} \quad \text { or } \quad \frac{1}{\mathrm{v}}=\frac{5-22.5}{22.5 \times 5} \quad \text { or } \quad \mathrm{v}=\frac{22.5 \times 5}{17.5}\\ \quad \text { or } \quad \mathrm{v}=-6.43 \mathrm{~cm}
The position of the final image =-(6.43-1.5) \mathrm{cm}=-4.93 \mathrm{~cm}

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