Question

An object is placed at 20cm from

the concave mirror whose radius

of curvature is 80cm. Calculate the

image distance and focal length of

the concave mirror.
​

Answer 1

Solution:-

Given

=> object distance ( u ) = 20 cm

=> Radius of curvature ( R ) = 80cm

Formula to find focal length ( f )

=> f = R/2

=> f = 80 / 2

=> f = 40 cm

As we know that in concave mirror focal length is always negative so f = - 40cm

Now we have to find image distance ( v )

Formula

=> 1/f = 1/v + 1/u

So using sign convention object distance is opposite to Incident ray so u = - 20 cm

Now

=> - 1/40 = 1/v - 1/20

=> 1/v = -1/40 + 1/20

=> 1/v = (- 1 + 2)/40

=> 1/v = 1 /40

=> v = 40 cm

So we get v = 40 cm so image form behind the mirror 40 cm from principal axis

=> Nature of image:- virtual and erect

Answer 2

Explanation:

Given :

⇒Tanθ = 20/21

To Find

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

First of all We have to find Sinθ and Cosθ

So take

⇒Tanθ = 20/21 = Perpendicular(p)/Base(b)

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = h

Using Pythagoras theorem

⇒h² = p² + b²

⇒h² = (20)² + (21)²

⇒h² = 400 + 441

⇒h² = 841

⇒h = √(841)

⇒h = 29

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = 29

Then

⇒Sinθ = P/h and Cosθ = b/h

⇒Sinθ = 20/29 and Cosθ  = 21/29

Now Put the value on

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

⇒(1-20/29 + 21/29)/(1+20/29 + 21/29)

⇒{(29-20+21)/29}/{29+20+21)/29}

⇒{(50 - 20)/29}/{(50+20)/29}

⇒(30/29)/(70/29)

⇒30/29 ×29/70

⇒30/70

⇒3/7

Answer = 3/7

Solution:-

Given

=> object distance ( u ) = 20 cm

=> Radius of curvature ( R ) = 80cm

Formula to find focal length ( f )

=> f = R/2

=> f = 80 / 2

=> f = 40 cm

As we know that in concave mirror focal length is always negative so f = - 40cm

Now we have to find image distance ( v )

Formula

=> 1/f = 1/v + 1/u

So using sign convention object distance is opposite to Incident ray so u = - 20 cm

Now

=> - 1/40 = 1/v - 1/20

=> 1/v = -1/40 + 1/20

=> 1/v = (- 1 + 2)/40

=> 1/v = 1 /40

=> v = 40 cm

So we get v = 40 cm so image form behind the mirror 40 cm from principal axis

=> Nature of image:- virtual and erect

hope this helps you!!

thank you ⭐