Question

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5ILL 01
I SO
11. When a concave mirror is placed facing the sun, the
sun's rays converge to a point 10 cm from the mirror.
Now, an erect, 2-cm-long pin is placed 15 cm away
on the principal axis of the mirror. If you want to get
the image of the pin on a card, where would you
place the card? What would be the nature and
height of the image?​

Answer 1

Answer:

It is given that rays are coming from sun. So we can consider the rays to come from a large distance or infinity in our case.

We know that, light rays traveling parallel to the principal axis and coming from infinity generally converge at the principal focus of the concave mirror.

Hence according to the given information, we can say that the principal focus lies 10 cm from the pole of the mirror.

Now we know that Curvature is nothing but 2 times of Focus. This means Curvature is at 2 × 10 = 20 cm.

Now according to the question the object is placed at 15 cm. The position of object lies between focus and curvature.

We know that object that lies between f and 2f will form an image beyond 2f.

Therefore the image of the pin will be formed beyond 20 cm. The nature of the image would be Real, inverted and Magnified image.

Now let's calculate the exact value with the help of Mirror's formula.

→ 1/v + 1/u = 1/f   ( Mirror's formula )

According to the question,

→ u = -15 cm, f = -10 cm, v = ?

Substituting the values we get,

→ 1/v - 1/15 = -1/10

→ 1/v = 1/15 - 1/10

→ 1/v = -1/30

→ v = -30 cm ( - sign represent sign convention )

Therefore the image is formed at 30 cm from the principal axis.

We know that,

m = -v/u = H/h

Here, H is the height of the image and h is the height of the object.

So substituting in the formula we get,

→ - [ -30/-15 ] = H/2

→ -2 = H/2

→ H = -4 cm [ - sign represents sign convention ]

Hence the height of the image is 4 cm below the principal axis (inverted).

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

• Focal Length (f) = -10 cm
• Object Distance (u) = -15 cm
• Hight of object (h) = 2 cm

Use mirror formula

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

⇒1/v = 1/f - 1/u

Put values

⇒1/v = 1/-10 -(1/-15)

⇒1/v = -1/10 + 1/15

⇒1/v = -3 + 2/30

⇒1/v = -1/30

⇒v = -30 cm

∴ Image distance = -30 cm

__________________________

As we know that :

Magnification = -v/u = h/h

⇒-(-30)/-15 = h/2

⇒30/-15 = h/2

⇒-2 = h/2

⇒h'= -4

∴ Height of Image is - 4 cm