Using the lens formula
"\\frac{1}{v}-\\frac1u=\\frac1f"
"\\frac1v-\\frac{1}{-4}=\\frac1{12}"
"v=-6cm"
here v is negative which means that the image formed is virtual
m="\\dfrac vu" ="\\dfrac IO"
"m=\\frac{-6}{-4}=\\frac{I}{1.2}\\\\I=1.8"
here value of I is positive which means that the image is upright
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