Question

Q43 The image formed by a convex mirror of focal length 20 cm is a quarter of the object. What is the distance of the object from the mirror? (a) -20 cm (b) - 40 cm (C) - 60 cm (d) -80 cm​

Answer 1

Answer:

Magnification of mirror = (-v)/u

According to the question, height of the image w.r.t the object is 0.25.

Therefore the magnification of the object is 0.25

⇒ -v/u = 0.25

⇒ -v = 0.25u

⇒ -4v = u

Also, it is given that the focal length of the mirror is 20 cm. Applying the mirror formula and calculating we get:

$\boxed{ \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}}\\\\\\\implies \dfrac{1}{20} = \dfrac{1}{-4v} + \dfrac{1}{v}\\\\\\\implies \dfrac{1}{20} = \dfrac{-1+4}{4v}\\\\\\\implies \dfrac{1}{20} = \dfrac{3}{4v}\\\\\\\implies 4v = 20 \times 3 = 60\\\\\implies \boxed{\bf{u = -4v = -60\:cm}}$

Hence Option (C) is the correct answer.

Answer 2

${\huge\underline{\underline{\text{Question:}}}}$

• The image formed by a convex mirror of focal length 20 cm is a quarter of the object. What is the distance of the object from the mirror?

(a) -20 cm

(b) - 40 cm

(C) - 60 cm

(d) -80 cm

.

${\huge\underline{\underline{\text{Solution:}}}}$

• $f = 20cm \: \: \: \text{...(given)}$

.

• $h \text{ is 4 times of } {h}^{′}. \: \: \: \text{...(given)}$

$\text{So We can say that,}$

• $h = 4 {h}^{′} \: \: \: ...(1)$

$\text{As we know that,}$

$\longrightarrow \boxed{m = \frac{{h}^{′}}{h}}$

$\longrightarrow m = \frac{{h}^{′}}{4{h}^{′}} \: \: \: \text{...(By [1])}$

$\longrightarrow \boxed{m = \frac{1}{4}}$

$\text{As we know that,}$

$\longrightarrow \boxed{m = \frac{ - v}{u}}$

$\text{So We can say that,}$

$\longrightarrow \frac{1}{4} = \frac{ - v}{u}$

$\longrightarrow u = - 4v \: \: \: ...(2)$

$\text{Mirror formula:}$

$\longrightarrow \boxed{ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} }$

$\longrightarrow { \frac{1}{20} = \frac{1}{ - 4v} + \frac{1}{v} } \: \: \: \text{...(By [2])}$

$\longrightarrow { \frac{1}{ 20} = - \frac{1}{ 4v} + \frac{1}{v} }$

$\longrightarrow { \frac{1}{ 20} = \frac{ - 1 + 4}{ 4v} }$

$\longrightarrow { \frac{1}{ 20} = \frac{3}{ 4v} }$

$\longrightarrow {4v}= {3(20)}$

$\longrightarrow {4v}= 60cm$

$\text{Substitute this value in Eq 2,}$

$\longrightarrow u = - 4v \: \: \: ...(2)$

$\longrightarrow u = - (60cm)$

$\longrightarrow \boxed{u = - 60cm}$

${\huge\underline{\underline{\text{Final \: Answer:}}}}$

• (C) - 60 cm