Question

An object forms a vertical image which is 1/8 of the size of the object. If the object is placed at a distance of 400 cm from the convex mirror. Calculate
i. position of the image
ii. focal length of the mirror
​

Answer 1

Correct question :-

An object forms a virtual image which is 1/8th of the size of the object. If the object is placed at a distance of 40 cm from the convex mirror, calculate

(i) the position of the image

(ii) the focal length of the mirror

Given :-

• Size of image $\sf (h_i)$ = 1/8 times of size of object $\sf (h_o)$

• Distance of object from convex mirror (u) = -40 cm

To Find :-

• Distance of image from convex mirror (v).

• Focal length of the mirror (f).

Solution :-

$\underline{\boxed{\bf Magnification , m = \dfrac{-v}{u}= \dfrac{h_i}{h_o}}}$

$\longrightarrow \sf \dfrac{-v}{-40} = \dfrac{\dfrac{1}{8} \times h_0}{h_0} \\\\\longrightarrow \dfrac{v}{40} = \dfrac{h_0}{8} \times \dfrac{1}{h_0} \\\\\longrightarrow \dfrac{v}{40} = \dfrac{1}{8} \\\\\longrightarrow v = \dfrac{40}{8} \\\\\longrightarrow \bf v = 5$

Distance of image from convex mirror = 5 cm

Using mirror formula,

$\underline{\boxed{\bf \dfrac{1}{f}= \dfrac{1}{u}+\dfrac{1}{v}}}$

$\longrightarrow \sf \dfrac{1}{f} = \dfrac{1}{-40}+\dfrac{1}{5} \\\\\longrightarrow \dfrac{1}{f} = \dfrac{5-40}{-40 \times 5 } \\\\\longrightarrow \dfrac{1}{f} = \dfrac{-35}{-200} \\\\\longrightarrow f = \dfrac{200}{35} \\\\\longrightarrow \bf f = 5.71$

Focal length of the mirror = 5.71 cm

Answer 2

Answer:

$Solution :- \\ </p><p></p><p>\\{\implies} \underline{\boxed{\bf Magnification , m = \dfrac{-v}{u}= \dfrac{h_i}{h_o}}}\\ </p><p></p><p> \begin{gathered}\longrightarrow \sf \dfrac{-v}{-40} \\ = \dfrac{\dfrac{1}{8} \times h_0}{h_0} \\\\\longrightarrow \\ \dfrac{v}{40} = \dfrac{h_0}{8} \times \dfrac{1}{h_0} \\ \\\\\longrightarrow \dfrac{v}{40} \\ = \dfrac{1}{8} \\\\\longrightarrow v = \dfrac{40}{8} \\ \\\\\longrightarrow \bf v = 5\end{gathered} \\ ⟶−40−v=h081×h0 \\ ⟶40v=8h0×h01 \\ ⟶40v=81 \\ ⟶v=840 \\ ⟶v=5</p><p></p><p>Distance of image from convex mirror = 5 cm \\ </p><p></p><p>Using mirror formula, \\ </p><p></p><p>\underline{\boxed{\bf \dfrac{1}{f}= \dfrac{1}{u}+\dfrac{1}{v}}}f1=u1+v1 \\ </p><p></p><p>\begin{gathered}\longrightarrow \sf \dfrac{1}{f} = \dfrac{1}{-40}+\dfrac{1}{5} \\\\\longrightarrow \dfrac{1}{f} = \dfrac{5-40}{-40 \times 5 } \\\\\longrightarrow \dfrac{1}{f} = \dfrac{-35}{-200} \\\\\longrightarrow f = \dfrac{200}{35} \\\\\longrightarrow \bf f = 5.71\end{gathered} \\ ⟶f1=−401+51 \\ ⟶f1=−40×55−40 \\ ⟶f1=−200−35 \\ ⟶f=35200 \\ ⟶f=5.71 \\ \\ </p><p></p><p>hence \: the \: focal \: lenth = 5.71$