Free Printable Light-Reflection And Refraction

1.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will you do first when you come across a numerical like this?

a)

b)

u = -4 cm (object distance which is negative as it is on the left side of the lens)

c)

f = -12 cm (focal length which is negative as it is on the left side of the lens)

d)

Lens formula : 1/v – 1/u = 1/f

e)

1/v – 1/-4 = 1/-12

2.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the object distance?

a)

1/v = - (1/12 ) – 1/4

b)

u = -4 cm (object distance which is negative as it is on the left side of the lens)

c)

f = -12 cm (focal length which is negative as it is on the left side of the lens)

d)

Lens formula : 1/v – 1/u = 1/f

e)

1/v – 1/-4 = 1/-12

3.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the focal length ?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3

c)

f = -12 cm (focal length which is negative as it is on the left side of the lens)

d)

Lens formula : 1/v – 1/u = 1/f

e)

1/v – 1/-4 = 1/-12

4.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the lens formula?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3

c)

V = - 3cm

d)

Lens formula : 1/v – 1/u = 1/f

e)

1/v – 1/-4 = 1/-12

5.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. On inserting the numerical values in the formula, what will the equation appear as?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3

c)

V = - 3cm

d)

The image is formed 3 cm in front of the concave lens and is smaller than the object.

e)

1/v – 1/-4 = 1/-12

6.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will be the third step of the numerical?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3

c)

V = - 3cm

d)

The image is formed 3 cm in front of the concave lens and is smaller than the object.

e)

Image is virtual and erect.

7.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will be the fourth step of the numerical?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3

c)

V = - 3cm

d)

The image is formed 3 cm in front of the concave lens and is smaller than the object.

e)

Image is virtual and erect.

8.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What will be the fourth step of the numerical?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3 ('u' is object distance, 'v' is image distance and 'f' is focal length)

c)

V = - 3cm ('u' is object distance, 'v' is image distance and 'f' is focal length)

d)

The image is formed 3 cm in front of the concave lens and is smaller than the object.

e)

Image is virtual and erect.

9.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the image distance?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3 ('u' is object distance, 'v' is image distance and 'f' is focal length)

c)

V = - 3cm ('u' is object distance, 'v' is image distance and 'f' is focal length)

d)

The image is formed 3 cm in front of the concave lens and is smaller than the object.

e)

Image is virtual and erect.

10.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the first characteristic of the image?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3 ('u' is object distance, 'v' is image distance and 'f' is focal length)

c)

V = - 3cm ('u' is object distance, 'v' is image distance and 'f' is focal length)

d)

The image is formed 3 cm in front of the concave lens and is smaller than the object.

e)

Image is virtual and erect.

11.

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image. What is the second characteristic of the image?

a)

1/v = - (1/12 ) – 1/4

b)

1/v = -4/12 = -1/3 ('u' is object distance, 'v' is image distance and 'f' is focal length)

c)

V = - 3cm ('u' is object distance, 'v' is image distance and 'f' is focal length)

d)

The image is formed 3 cm in front of the concave lens and is smaller than the object.

e)

Image is virtual and erect.

12.

A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either? What is the first statement?

a)

: The focal length of a mirror does not depend upon the nature of the medium in which it is placed

b)

whereas the focal length of a lens depends upon the medium in which it is placed.

c)

Thus, there will be no change in the focal length of the concave mirror whereas the focal length of the convex lens will change.

d)

If the refractive index of the material of the convex lens is greater than water it will increase,

e)

if it is equal to water it will become infinity and if it is less than water it will be negative.

13.

A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either? What is the Second statement?

a)

: The focal length of a mirror does not depend upon the nature of the medium in which it is placed

b)

whereas the focal length of a lens depends upon the medium in which it is placed.

c)

Thus, there will be no change in the focal length of the concave mirror whereas the focal length of the convex lens will change.

d)

If the refractive index of the material of the convex lens is greater than water it will increase,

e)

if it is equal to water it will become infinity and if it is less than water it will be negative.

14.

A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either? What is the third statement?

a)

: The focal length of a mirror does not depend upon the nature of the medium in which it is placed

b)

whereas the focal length of a lens depends upon the medium in which it is placed.

c)

Thus, there will be no change in the focal length of the concave mirror whereas the focal length of the convex lens will change.

d)

If the refractive index of the material of the convex lens is greater than water it will increase,

e)

if it is equal to water it will become infinity and if it is less than water it will be negative.

15.

A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either? What is the Fourth statement?

a)

: The focal length of a mirror does not depend upon the nature of the medium in which it is placed

b)

whereas the focal length of a lens depends upon the medium in which it is placed.

c)

Thus, there will be no change in the focal length of the concave mirror whereas the focal length of the convex lens will change.

d)

If the refractive index of the material of the convex lens is greater than water it will increase,

e)

if it is equal to water it will become infinity and if it is less than water it will be negative.

16.

A concave mirror and a convex lens are held separately in water. What changes, if any, do you expect in the focal length of either? What is the Fifth statement?

a)

: The focal length of a mirror does not depend upon the nature of the medium in which it is placed

b)

whereas the focal length of a lens depends upon the medium in which it is placed.

c)

Thus, there will be no change in the focal length of the concave mirror whereas the focal length of the convex lens will change.

d)

If the refractive index of the material of the convex lens is greater than water it will increase,

e)

if it is equal to water it will become infinity and if it is less than water it will be negative.

17.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What is the first given information that you will use?

a)

distance of the object (u) = - 20 cm

b)

Height of the object (h1 ) = 5 cm,

c)

focal length (f) = 10 cm,

d)

Image distance (v) = ?

e)

Height of the image (h2 ) = ?

18.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What is the Second given information that you will use?

a)

1/v - 1/u = 1/f

b)

Height of the object (h1 ) = 5 cm,

c)

focal length (f) = 10 cm,

d)

Image distance (v) = ?

e)

Height of the image (h2 ) = ?

19.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What is the Third given information that you will use?

a)

1/v - 1/u = 1/f

b)

$\frac{1}{v}-\frac{1}{\left(-20\right)}=\ \frac{1}{10}$

c)

focal length (f) = 10 cm,

d)

Image distance (v) = ?

e)

Height of the image (h2 ) = ?

20.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What is the first thing you are supposed to find?

a)

1/v - 1/u = 1/f

b)

$\frac{1}{v}-\frac{1}{\left(-20\right)}=\ \frac{1}{10}$

c)

focal length (f) = 10 cm,

d)

Image distance (v) = ?

e)

Height of the image (h2 ) = ?

21.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What is the Second thing you are supposed to find?

a)

1/v - 1/u = 1/f

b)

$\frac{1}{v}-\frac{1}{\left(-20\right)}=\ \frac{1}{10}$

c)

$\frac{1}{v}=\frac{1}{10}+\frac{1}{\left(-20\right)}$

d)

Image distance (v) = ?

e)

Height of the image (h2 ) = ?

22.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What is the third thing you are supposed to find?

a)

1/v - 1/u = 1/f

b)

$\frac{1}{v}-\frac{1}{\left(-20\right)}=\ \frac{1}{10}$

c)

$\frac{1}{v}=\frac{1}{10}+\frac{1}{\left(-20\right)}$

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

Magnification (M) = ?

23.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What is the lense formula for convex lens?

a)

1/v - 1/u = 1/f

b)

$\frac{1}{v}-\frac{1}{\left(-20\right)}=\ \frac{1}{10}$

c)

$\frac{1}{v}=\frac{1}{10}+\frac{1}{\left(-20\right)}$

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

Magnification (M) = ?

24.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What will it look like when the numerical values are inserted

a)

$\frac{1}{v\ \ }=\ \frac{20-10}{200}$

b)

$\frac{1}{v}-\frac{1}{\left(-20\right)}=\ \frac{1}{10}$

c)

$\frac{1}{v}=\frac{1}{10}+\frac{1}{\left(-20\right)}$

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

25.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What will be the next step after the the numerical values are inserted

a)

$\frac{1}{v\ \ }=\ \frac{20-10}{200}$

b)

$\frac{1}{v}-\frac{1}{\left(-20\right)}=\ \frac{1}{10}$

c)

$\frac{1}{v}=\frac{1}{10}+\frac{1}{\left(-20\right)}$

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

26.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What will be the third step after the the numerical values are inserted

a)

$\frac{1}{v\ \ }=\ \frac{20-10}{200}$

b)

$\frac{1}{v\ }=\ \frac{10}{200}$

c)

$\frac{1}{v}=\frac{1}{10}+\frac{1}{\left(-20\right)}$

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

27.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? What will be the Fourth step after the the numerical values are inserted

a)

$\frac{1}{v\ \ }=\ \frac{20-10}{200}$

b)

$\frac{1}{v\ }=\ \frac{10}{200}$

c)

v = 20

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

28.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? how will the equation look when the denominators are equalised?

a)

$\frac{1}{v\ \ }=\ \frac{20-10}{200}$

b)

$\frac{1}{v\ }=\ \frac{10}{200}$

c)

v = 20

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

29.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? how will the equation look when the denominators are equalised and subtraction is complete?

a)

$\frac{1}{v\ \ }=\ \frac{20-10}{200}$

b)

$\frac{1}{v\ }=\ \frac{10}{200}$

c)

v = 20

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

30.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object? what is the image distance?

a)

$\frac{1}{v\ \ }=\ \frac{20-10}{200}$

b)

The positive sign of the image distance shows that image is formed at 20 cm on the right side of the lens.

c)

v = 20

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

31.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? how will you calculate magnification?

a)

$\frac{h1}{h2}=\frac{v}{u}$

b)

The positive sign of the image distance shows that image is formed at 20 cm on the right side of the lens.

c)

v = 20

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

32.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? what is the nature of the image?

a)

$\frac{h1}{h2}=\frac{v}{u}$

b)

The positive sign of the image distance shows that image is formed at 20 cm on the right side of the lens.

c)

$h2\ =\ \frac{v}{u\ }x\ h1$

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

33.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? What is the value of h2

a)

$\frac{h1}{h2}=\frac{v}{u}$

b)

$h2\ =\ \frac{20}{\left(-20\right)}x\ 5$

c)

$h2\ =\ \frac{v}{u\ }x\ h1$

d)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

34.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? What is the value of h2 on inserting numerical value?

a)

$\frac{h1}{h2}=\frac{v}{u}$

b)

$h2\ =\ \frac{20}{\left(-20\right)}x\ 5$

c)

$h2\ =\ \frac{v}{u\ }x\ h1$

d)

$h2\ =\ -5$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

35.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? What is the numerical value of h2?

a)

$\frac{h1}{h2}=\frac{v}{u}$

b)

$h2\ =\ \frac{20}{\left(-20\right)}x\ 5$

c)

$h2\ =\ \frac{v}{u\ }x\ h1$

d)

$h2\ =\ -5$

e)

$\frac{1}{v\ }=\ \frac{1}{10}-\frac{1}{20}$

36.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? What is Mafnification

a)

Magnification : $\frac{20}{-20}=\ -1$

b)

$h2\ =\ \frac{20}{\left(-20\right)}x\ 5$

c)

$h2\ =\ \frac{v}{u\ }x\ h1$

d)

$h2\ =\ -5$

e)

The image is inverted because h2 is negetive.

37.

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? What will you interfret from negetive magnification and negetive h2

a)

Magnification : $\frac{20}{-20}=\ -1$

b)

The negative sign of the height of the image and the magnification shows that the image is inverted and real. It is below the principal axis and is of the same size as the object.

c)

$h2\ =\ \frac{v}{u\ }x\ h1$

d)

$h2\ =\ -5$

e)

The image is inverted because h2 is negetive.

Light-Reflection And Refraction_7

Total questions: 37

Worksheet time: 19mins

Similar Resources on Wayground