Question

An object of length 4 cm is placed in front of a concave mirror at a distance 30 cm. The focal length
of the mirror is 15 cm.
i) Where will the image form?
ii) What will be the length of the image?​​

Answer 1

Given :

• Height of image, $\sf h_{i}$ = 4 cm
• Object distance, u = - 30 cm
• Focal length, f = - 15 cm

⠀⠀⠀⠀

To find :

• Where will the image form?
• What will be the length of the image?

⠀⠀⠀⠀

Solution :

⠀⠀⠀⠀

$\bigstar\:{\underline{\sf Using\:Mirror\:Formula\::}}$

$\star\;{\boxed{\sf{\pink{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}$

$:\implies\sf \dfrac{1}{-15} = \dfrac{1}{v} + \dfrac{1}{-30}$

$:\implies\sf \dfrac{1}{v} = \dfrac{1}{-15} + \dfrac{1}{30}$

$:\implies\sf \dfrac{1}{v} = \dfrac{- 2 + 1}{30}$

$:\implies\sf \dfrac{1}{v} = \dfrac{- 1}{30}$

$:\implies{\underline{\boxed{\sf{\purple{v = {\frak{-30\:cm}}}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Distance\;of\;image\;from\:is\; \bf{30\:cm}.}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━

$\bigstar\:{\underline{\sf Now,\: from\:Magnification\:Formula\::}}$

$\star\;{\boxed{\sf{\pink{m = \dfrac{h_i}{h_o} = \dfrac{-v}{u}}}}}$

$:\implies\sf \dfrac{h_i}{4} = \dfrac{- \cancel{(-30)}}{ \cancel{-30}}$

$:\implies\sf \dfrac{h_i}{4} = - 1$

$:\implies\sf h_i = -1 \times 4$

$:\implies{\underline{\boxed{\sf{\purple{h_i = {\frak{-4\:cm}}}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Length\;of\;image\;is\; \bf{4\:cm}.}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━

$\bigstar\:{\underline{\sf Nature\:of\:image\::}}$

At centre of curvature (C).

Real and inverted.

Answer 2

$\green{\mid{\fbox{\tt{ ❝Aɴꜱᴡᴇʀ੭❞ }}\mid}}$

The relationship between the radius of curvature and focal length is that the radius of curvature is twice the focal length.

In this case, focal length of the concave mirror is given as 15 cm. Hence, the radius of curvature is 15 x 2 = 30 cm.

When the object is placed on the centre of curvature, on the principal axis, in front of a concave lens, the image is formed at the same point. The image formed is real and inverted.

So, in this case, the object distance is equal to the image distance. Therefore, the object size is equal to the image size.

As the object is of 4 cm in length, the image will also be of 4 cm in length.

━━━━━━━━━━━━━━━━━━━━

$\purple{\mid{\fbox{\tt{ ❝Tʜᴀɴᴋ Yᴏᴜ❞ }}\mid}}$