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Question Number 133115 by aurpeyz last updated on 19/Feb/21
an object is placed 5cm away from a  plane mirror. the mirror is rotated  tbrough angle 20^0  about the point of  incidence. calculate the shortest  distance between the old and the new  position of the image
$${an}\:{object}\:{is}\:{placed}\:\mathrm{5}{cm}\:{away}\:{from}\:{a} \\ $$$${plane}\:{mirror}.\:{the}\:{mirror}\:{is}\:{rotated} \\ $$$${tbrough}\:{angle}\:\mathrm{20}^{\mathrm{0}} \:{about}\:{the}\:{point}\:{of} \\ $$$${incidence}.\:{calculate}\:{the}\:{shortest} \\ $$$${distance}\:{between}\:{the}\:{old}\:{and}\:{the}\:{new} \\ $$$${position}\:{of}\:{the}\:{image} \\ $$
Answered by mr W last updated on 19/Feb/21
Commented by mr W last updated on 19/Feb/21
A=object  blue=mirror before  red=mirror after rotation  B=image before  C=image after rotation of mirror  a=distance of object to mirror before  b=distance of object to mirror after  c=distance new image to old image    c=2a sin θ=2×5×sin 20°≈3.42 cm
$${A}={object} \\ $$$${blue}={mirror}\:{before} \\ $$$${red}={mirror}\:{after}\:{rotation} \\ $$$${B}={image}\:{before} \\ $$$${C}={image}\:{after}\:{rotation}\:{of}\:{mirror} \\ $$$${a}={distance}\:{of}\:{object}\:{to}\:{mirror}\:{before} \\ $$$${b}={distance}\:{of}\:{object}\:{to}\:{mirror}\:{after} \\ $$$${c}={distance}\:{new}\:{image}\:{to}\:{old}\:{image} \\ $$$$ \\ $$$${c}=\mathrm{2}{a}\:\mathrm{sin}\:\theta=\mathrm{2}×\mathrm{5}×\mathrm{sin}\:\mathrm{20}°\approx\mathrm{3}.\mathrm{42}\:{cm} \\ $$
Commented by aurpeyz last updated on 19/Feb/21
i am very happy. thank you Sir
$${i}\:{am}\:{very}\:{happy}.\:{thank}\:{you}\:{Sir} \\ $$

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