Question

1. An object is placed 6cm in front of a concave mirror of Radius of curvature 24cm. Find the position, nature and size of the image?
(a)v=-12cm, hi=-8cm, real
(b)v=8cm, hi=12cm,virtual
(c)v=12cm, hi=8cm, virtual
(d)v=6cm, hi=12cm, virtual
2. An object 3cm high is placed at a distance of 16cm from concave mirror which produces a real image 3cm high.
(i) Find position of the image?
(ii)Find focal length of mirror?
(a)(i)-24cm (ii)- 9.6cm (b)(i) -16cm (ii)-8.5cm (c)(i)-24cm (ii)-8.5cm (d) (i)24cm (ii)-9.6 cm
3. Magnification of image formed by convex mirror of focal length 15cm is 1/3. Find the object distance?
(a)-15cm (b)-45cm (c)-20cm (d)-30cm
4. If an object is placed 10cm from a convex mirror of radius of curvature 60cm, then find the position of the image?
(a)8.5cm (b)7.5cm (c)12.5cm (d)10cm
5. A 4.5cm needle is placed 12cm away from a convex mirror of focal length 15cm. Give the location of image and magnification?
(a)v=11.6cm, m=3/2 (b) v=7.5cm, m=0.75 (c)v=6.6cm, m=0.55 (d)v= 9.25cm, m=0.25
PLEASE HELP ME FAST!!!!

Answer 1

Answer:

first answer is (a) and second is

Answer 2

Answer:

1. option c 2. option b 3. option d 4. option b 5. option c

Explanation:

1. object distance (u)= -6cm

Radius of curvature (R)= -24 cm

Focus of the mirror is (f)=-24/2= -12 cm

Mirror formula  is given by   $\frac{1}{v} +\frac{1}{u} =\frac{1}{f}$,

Putting the values in the formula we get image distance (v)=$\frac{1}{v} =\frac{1}{f} -\frac{1}{u} =\frac{1}{12}$

⇒ v=12 cm, It is positive hence it is virtual,$\frac{h_i}{h_0}=\frac{-v}{u}=2$.

option (C) is correct.

2. object height = 3cm

object distance = -16cm

real image height= 3 cm

Using the formula for magnification $\frac{h_i}{h_0}=3/3=1\\implies \frac{-v}{u}=1 \implies v=-16cm$

Using the mirror formula we get $\frac{1}{f} =\frac{1}{-16} +\frac{1}{-16}=\frac{-1}{8} \\\implies f= -8cm$

option (b) is correct.

3. magnification by convex mirror (m)$m = \frac{-v}{u}=\frac{1}{3}\implies v=\frac{-u}{3}$

focal length = 15 cm

using the mirror formula we get $\frac{1}{v} +\frac{1}{u}=\frac{1}{15}$, putting the value of u from the above equation we get $\frac{-3}{u} +\frac{1}{u} = \frac{1}{15} \implies u =-30 cm$

option(d ) is correct

4. object distance (u)= -10cm

radius of curvature = 60cm

focal length = 60/2 = 30 cm

image distance can be found from mirror formula

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{30}-\frac{-1}{10}\\\implies v=7.5 cm$

option (b) is correct

5. height of needle = 4.5 cm

distance of needle from convex mirror (u)= - 12 cm

focal length = 15 cm

from the mirror formula we can get image distance

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{-1}{12}\\\implies v=6.6 cm$

$m = \frac{-v}{u} = \frac{-6.6}{-12}= 0.55$

option(c) is correct.