Ray Optics and Optical Instruments Test - 11

Let the radius of curvature of the convex lens are denoted as
For a convex lens
R₁ =+40cm [As it is towards right]
R₂ =-40cm [As it is towards left]
The refractive index of the lens [µ]=1.5
Let f is the focal length of the lens.
From lens maker 's formula we know that –
1/f=[µ-1] [1/R₁ -1/R₂ ]
1/f=[1.5-1][1/40-(1/-40)]
1/f=0.5 x (2/40)
1/f=1/40
f=40cm
Hence, the focal length of the lens is 40 cm.

Given, u=−20cm
m=−3 ⇒−v/u ​=−3
v=−60cm
1/f=(1/v ​)+(1/u)​=(-1/60)​−(1/20)​=−1 −3/60 ​
f=−60/4 ​=−15cm
 

The answer is of the same size because as the focal length of the concave mirror is 15cm then the radius of curvature becomes 30 cm and the object will be on center of curvature by which image will form the same on center of curvature which is inverted and of same length.

-Virtual image cannot be formed from a virtual object.
- When an object is placed between pole and focus, the image formed by the concave mirror is magnified, virtual and erect.
-When an object is placed beyond the centre of curvature , the image of it is formed between the centre of curvature and focus which is diminished,real and inverted,so the real image of the real object can be formed by a concave mirror.
-When the object is virtual, the image formed is real for the concave surface as shown in figure.
 

We are given that:
Focal length = 20 cm
In a convex mirror image is formed between focal length and mirror.
So the maximum distance is f i.e. focal length of the mirror.  
Now
f = 20 / 2 = 10 cm
Thus the largest distance of an image from a convex mirror is 10 cm.
 

1/v + 1/u = 1/f
⇒(-1/v2)x(dv/dt)-(1/u2)(du/dt)=0
dv/dt=-(v2/u2)(du/dt)
For spherical mirror: u is negative and particle is moving towards mirror i.e. u is increasing and du/dt is positive, therefore dv/dt must be negative or v should decrease.
Case-1 ⇒Image real ⇒v negative ⇒decreasing v means image moving away from mirror
Case-2 ⇒Image virtual ⇒v positive ⇒decreasing v means image moving towards mirror
So, image may move towards the mirror  
 

Any line joining the object point and the image point always passes through the centre of curvature of the mirror. This is because of the law, that any ray passing through the centre of curvature of the mirror will retrace its path back.
Thus, any ray passing through the centre of curvature will be perpendicular to the mirror irrespective of the type of the mirror used.

Focal length is the property of a spherical mirror. It does not depend on the light used. So, it is the same for all lights.

Concave mirror produces a virtual image larger than the object when the object is placed  between the centre of curvature and the pole of the mirror.
Convex lens can produce a virtual and enlarged image when the object is placed between the pole and focus of the lens.

When an object is placed at the centre of curvature then image is formed at radius of curvature. 
So, the minimum distance is zero.
OR
Let focal length be f and an object is placed at u=2f
1/−f=(1/v)+(1/u)=(1/v)+(1/−2f)
v=−2f. 
Image will form at the object location. 
So the minimum distance is zero.