Locate the image formed by refraction in the situation shown in figure. The point C is the centre of curvature.
Given,
μ1=1μ2=1.5R=20 cm(Radius of curvature)u=−25 cm\mu_1=1\\\mu_2=1.5\\R=20\ cm(\text{Radius of curvature})\\u=-25\ cmμ1=1μ2=1.5R=20 cm(Radius of curvature)u=−25 cm
So, Using Formula :
⇒μ2v−μ1u=μ2−μ1R ⇒1.5v=0.520−125\Rightarrow \dfrac{\mu_2}{v}-\dfrac{\mu_1}{u}=\dfrac{\mu_2-\mu_1}{R}\\\ \\\Rightarrow \dfrac{1.5}{v}=\dfrac{0.5}{20}-\dfrac{1}{25}⇒vμ2−uμ1=Rμ2−μ1 ⇒v1.5=200.5−251
⇒1.5v=−3200⇒v=−200×0.5=−100 cm\Rightarrow \dfrac{1.5}{v}=\dfrac{-3}{200}\\\Rightarrow v=-200\times0.5=-100\ cm⇒v1.5=200−3⇒v=−200×0.5=−100 cm
Answer:
So, the image is 100cm from (P) and on the surface on the side of S.
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