Question

33. The image of an object is formed on a screen at 1 1/2 of the distance of the object using a
concave mirror of focal length 9 cm. Find the object distance, image distance and
magnification​

Answer 1

Answer:

• 14.14 object distance
• Magnification = -0.38

Explanation:

Given:-

• Image distance ,u = -11/2 = -5.5 (as image is real).
• Focal length ,f = -9 cm

To Find:-

• Object Distance & Magnification

Solution:-

As we have to calculate the object distance . Using mirror formula .

• 1/v + 1/u = 1/f

where,

u is the object distance

f is the focal length

v is the image distance

Substitute the value we get

→ 1/(- 5.5) + 1/u = 1/(-9)

→ -1/u = -1/9 + 1/5.5

→ -1/u = -1/9 + 10/55

→ -1/u = -55+90/495

→ -1/u = 35/495

→ u = -495/35

→ u = -14.14 cm

Hence ,the object distance is 14.14

Now, calculating the Magnification

• m = -v/u

where

v image distance

u object distance

Substitute the value we get

→ m = -(-5.5/-14.14)

→ m = -5.5/14.14

→ m = -0.33

Hence , the Magnification of the mirror is -0.38

Answer 2

Answer:

Given :-

• Image distance = -5.5
• Focal length = 9.5 cm

To Find :-

• Object Distance
• Magnification

Solution :-

${\huge {\mathfrak {\red { \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}}}$

Here,

F indicate Focal length

V indicates Image distance

U indicate Object distance

$\tt \: \dfrac{1}{ - 5.5} + \dfrac{1}{u} \ = \dfrac{1}{-9}$

$\tt \: \dfrac{ - 1}{u} = \dfrac{-1}{9} + \dfrac{1}{5.5}$

$\tt \dfrac{ - 1}{u} = \dfrac{ - 1}{9} + \dfrac{10}{55}$

$\tt \dfrac{ - 1}{u} = \dfrac{ - 90 + 55}{495}$

$\tt \dfrac{ - 1}{u} = \dfrac{35}{495}$

$\tt \: u = - 14.14$

Hence, The object distance is 14.14 cm.

$\huge \bf \: m = \frac{ - v}{u}$

Magnification = -(-5.5)/-14.14

Magnification = -5.5/14.14

Hence, Magnification of mirror is 0.33.