
The image formed by convex mirror of focal length $30 \mathrm{cm}$ is the quarter of size of
the object. Then the distance of the object from the mirror is,
A. $30 \mathrm{cm}$
B. $90 \mathrm{cm}$
C. $120 \mathrm{cm}$
D. $60 \mathrm{cm}$
Answer
180k+ views
Hint Optics is the part of material science that reviews the conduct and properties of light, incorporating its communications with issues and the development of instruments that utilize or distinguish it. Optics as a rule portrays the conduct of obvious, bright, and infrared light. We must keep this in mind before attempting any such question. To proceed we must put the formula for calculation of focal length and formula for magnification.
Complete step by step answer:
From the given question, we know that, the focal length of the convex mirror is $\mathrm{f}=30 \mathrm{cm}$ and that the magnification of the image is $\text{m}=\dfrac{1}{4}$. We have to calculate the object distance that is denoted by u. According to the sign convention, we know that the measurements along the direction of light are taken as positive and that opposite to the light are taken as negative. Therefore, we derive that the transverse measurement above the principal axis is taken as positive and that below the principal axis is taken as negative. Therefore, $f=+30 \mathrm{cm}$ and $\mathrm{m}=+1 / 4=+0.25$.
Using the formula, $\mathrm{m}=\dfrac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}$
$\Rightarrow \text{mf}-\text{mu}=\text{f}$
$\Rightarrow \text{u}=\dfrac{\text{mf}-\text{f}}{\text{m}}=\dfrac{\text{m}-1}{\text{m}}\text{f}=\dfrac{0.25-1}{0.25}(30)=-90\text{cm}$
The minus sign prevails due to the assumption in direction but because we are looking into magnitudes only.
The correct answer is Option B.
Note: We must keep in mind that in optics, the sign related with the result portrays the direction only but to answer such questions, we only consider magnitudes.
Complete step by step answer:
From the given question, we know that, the focal length of the convex mirror is $\mathrm{f}=30 \mathrm{cm}$ and that the magnification of the image is $\text{m}=\dfrac{1}{4}$. We have to calculate the object distance that is denoted by u. According to the sign convention, we know that the measurements along the direction of light are taken as positive and that opposite to the light are taken as negative. Therefore, we derive that the transverse measurement above the principal axis is taken as positive and that below the principal axis is taken as negative. Therefore, $f=+30 \mathrm{cm}$ and $\mathrm{m}=+1 / 4=+0.25$.
Using the formula, $\mathrm{m}=\dfrac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}$
$\Rightarrow \text{mf}-\text{mu}=\text{f}$
$\Rightarrow \text{u}=\dfrac{\text{mf}-\text{f}}{\text{m}}=\dfrac{\text{m}-1}{\text{m}}\text{f}=\dfrac{0.25-1}{0.25}(30)=-90\text{cm}$
The minus sign prevails due to the assumption in direction but because we are looking into magnitudes only.
The correct answer is Option B.
Note: We must keep in mind that in optics, the sign related with the result portrays the direction only but to answer such questions, we only consider magnitudes.
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