Question

An object 5cm is placed at a distance of 40cm in front of convex mirror of radius of curvature 30cm. Find the position of image, size and nature of image.​

Answer 1

$\LARGE\underbrace{\purple{\sf{Required \; Solution:}}}$

Position of Image: Behind the mirror.  

Nature of Image: Virtual and Erect.  

Size of the Image: Diminished.

Explanation:  

Given:  

• Height of object = 5 cm  

• Object Distance (u) = - 20 cm (By using Sign Convention)  

• Radius of Curvature (R) = 30 cm  

• Focal Length (f) = 2R = 30 / 2 = 15 cm (By using Sign Convention)

$\underline{\frak{\star \; Using \; Mirror \; Formula:}}$

${\bigstar \; \underline{\boxed{\frak{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}}}}$

${\implies \sf{\dfrac{1}{v} - \dfrac{1}{20} = \dfrac{1}{15}}}\\\\{\implies \sf{\dfrac{1}{v}= \dfrac{1}{15} + \dfrac{1}{20}}}\\\\{\implies \sf{\dfrac{1}{v}= 4 + \dfrac{3}{60} \; \; (Taking \; LCM \; of \; 15 ;\ \& \; 20)}}\\\\{\implies \sf{\dfrac{1}{v} = \dfrac{7}{60}}}$

Image distance is 8.57 cm behind the convex mirror. As the v is positive the image formed is virtual and erect.

${\bigstar \; \underline{\boxed{\frak{Magnification = -\dfrac{v}{u}}}}}$

$\sf{= \dfrac{- 8.57 }{- 20 }= 0.4 cm}$

This indicates that the image is diminished because the magnification of the image is less than 1.

${\bigstar \; \underline{\boxed{\frak{Magnification = \dfrac{Height \; of \; Object}{ Height \; of \; image}}}}}$

$\sf{= \dfrac{hi}{ho} = 0.4}$

We know ho here,

$\sf{\dfrac{hi }{5} = 0.4}$

$\sf{hi = 0.4 \times 5}$  

${\underline{\boxed{\frak{\purple{hi = 2 cm}}}} \; \bigstar}$

 

As the image formed is smaller than the height of the object the image formed is diminished.

Therefore the final answer,  

Position of Image: Behind the mirror.  

Nature of Image: Virtual and Erect.  

Size of the Image: Diminished.

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