## Refraction Of Light At Spherical Surface Lens Multiple Choice Questions

**Question 1. What sort of lens will an air bubble in water behave like?**

- Biconvex
- Concavo-convex
- Biconcave
- Convexo-concave

**Answer:** 4. Convexo-concave

**Question 2. Observe the behaviour of the light rays as shown in the relation of n _{1} and n_{2} is**

- n
_{2}>n_{1} - n
_{1 }>>n_{2} - n > n
- n = n

**Answer:**

**Question 3. An object behaves like a convex lens in air and a concave lens in water. The refractive index of the material of the object is**

- Less than air
- More than both water and air
- More than air but less than water
- Almost equal to water

**Answer:** 3. More than air but less than water

**Read and Learn More Class 12 Physics Multiple Choice Questions**

**Question 4. The optical centre of a lens is a fixed point whose position is**

- Within the lens
- Outside the lens
- On the principal axis of the lens
- At the focus of the lens

**Answer:** 3. On the principal axis of the lens

**Question 5. A convex lens of focal length 20 cm is placed on a plane mirror. A point object is placed at a distance of 20 cm above the lens along its axis. What will be die final image distance from the lens?**

- 10 cm
- 1
- Infinity
- 20 cm
- 0

**Answer:** 3. 20 cm

**Question 6. If an object is placed at the focus of a concave lens, the image will be formed**

- At infinity
- At the mid-point of the optical centre and the focus
- At the optical centre
- At the focus

**Answer:** 2. At the mid-point of the optical centre and the focus

**Question 7. The focal length of a convex lens is if an object is placed at a distance u from the lens, the condition of formation of an inverted image of equal size as the object is**

- u = 2f
- u>2f
- f<u<2f
- 0 <u<f

**Answer:** 1. u = 2f

**Question 8. The focal length of a convex lens is If an object is placed at a distance u from the lens, the condition of formation of a diminished inverted image is**

- u = 2f
- u>2f
- f< u < 2f
- 0 < u <f

**Answer:** 2. u>2f

**Question 9. The focal length of a convex lens. If an object be placed at a distance u from the lens, the condition of formation of a virtual image is**

- u = f
- u > 2f
- f< u < 2f
- 0 <u<f

**Answer:** 4. 0 <u<f

**Question 10. The focal length of a convex lens.If an object is placed at a distance u from the lens, the condition of formation of an image at infinity is**

- u = f
- u>2f
- f< u < 2f
- 0 <u<f

**Answer:** 1. u = f

**Question 11. The focal length of a convex lens. If an object be placed at a distance u from the lens, the condition of formation of an inverted magnified image is**

- u=f
- u > 2f
- f< u < 2f
- 0 < u<f

**Answer:** 3. f< u < 2f

**Question 12. The focal length of a convex lens. If an object is placed at a distance u from the lens, the condition of formation of a magnified virtual image is**

- u = f
- u>2fF
- f< u< 2f
- 0<u<f

**Answer:** 4. 0<u<f

**Question 13. The focal length of a concave lens is f. If an object is placed at a distance u from the lens, the condition of formation of a diminished image is**

- u = 0
- 0 < u < oo
- u < 0 and |u| <|f|
- A diminished image will not be formed under any condition

**Answer:** 2. 0 < u < ∞

**Question 14. The focal length of a concave lens is f. If an object is placed at a distance u from the lens, the condition of formation of a real image is**

- u = 0
- 0 < u < oo
- u < 0 and| u| <|f|
- Real image will not be formed under any condition

**Answer:** 3. u < 0 and| u| <|f|

**Question 15. A point object is placed at the centre of a glass sphere. If the radius of the sphere is 6 cm and the refractive index of the material is 1.5, then the distance of the virtual image from the surface of the sphere will be**

- 2cm
- 4 cm
- 6 cm
- 12 cm

**Answer:** 3. 6 m

**Question 16. An object is placed at a distance of 20 cm from a convex lens of focal length 0 cm. The image distance is**

- 20 cm
- 6.67 cm
- 10 cm
- 30 cm

**Answer:** 1. 20 cm

**Question 17. The size of the image of an object which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance at 26 cm from the convex lens, the real size of the image would be**

- 1.25 cm
- 2.5 cm
- 1.05 cm
- 2 cm

**Answer:** 2. 2.5 cm

**Question 18. A convex lens image of focal of a length object.30 cm produces 5 times the magnified real image of an object, What is the object distance?**

- 36 cm
- 25cm
- 30cm
- 150 cm

**Answer:** 1. 36 cm

**Question 19. If the object distance from the first focus of a convex lens is x and the image distance from the second focus of the lens is x’, the graph between x and x’ will be a**

- Hyperbola
- Parabola
- Circle
- Ellipse

**Answer:** 1. Hyperbola

**Question 20. The focal length of a lens made of glass in air is 10 cm. What will be the focal length of the lens in water? The refractive index of glass =1.51 and the refractive index of water = 1.33.**

**Answer:**

- 18.84 cm
- 36 cm
- 18 cm
- 37.7 cm

**Answer:** 4. 37.7 cm

**Question 21. If a lens is surrounded by a medium denser than air, the focal length of the lens**

- Decreases
- 1ncreases
- Remains same
- Cannot be determined

**Answer:** 2. Increases

**Question 22. If the focal length of a symmetrical convex lens is equal to the radius of curvature of the lens, the refractive index of the material of the lens is**

- 0.5
- 1.1
- -1.5
- 1.5

**Answer:** 4. 1.5

**Question 23. The refractive index of the material of a double equi-convex lens is 1.5. If R be its radius of curvature then its focal length is**

- 0
- 2R
- R1
- R

**Answer:** 4. R

**Question 24. What type of lens is used in sunglasses?**

- A concavo-convex lens whose radii of curvature of the two surfaces are equal
- A biconcave lens whose radii of curvature of the two surfaces are equal
- A biconcave lens whose radii of curvature of the two surfaces are unequal
- Plano-concave lens

**Answer: **1. A concavo-convex lens whose radii of curvature of the two surfaces are equal

**Question 25. An equal-convex lens is divided into two halves along) XOX’ and (ii) YOY’ as shown in the. Suppose the focal lengths of the complete lens, of each half portion of **

**Case 1- ****And of each half portion of the
**

**Case 2 –**

**Respectively in this case the correct statement is**

- f = 2f; f”= f
- f = 2f; f”= f
- f = 2f; f”= 2f
- f = f; f”= f

**Answer:** 4. f = f; f”= f

**Question 26. A convex lens is immersed in a liquid whose refractive index is equal to that of the material of the lens. Under this condition the focal length of the lens**

**Answer:**

- Will decrease but will not be zero
- Will remain unchanged
- Will be zero
- Will be of infinite value

**Answer:** 4. Will be of infinite value

**Question 27. A convex lens made of glass has a focal length of 0.15 m in the air. if the refractive indices of glass and water are respectively \(\frac{3}{2}\) and \(\frac{4}{3}\), then the focal length of the lens immersed in water will be**

- 0.45 m
- 0.15 m
- 0.30 m
- 0.6 m

**Answer:** 4. 0.6 m

**Question 28. Which of the following is true for rays coming from infinity incident on the lens**

- Two images are formed.
- A continuous image is formed between the focal points of the upper and lower lens
- One image is formed
- None of the above

**Answer:** 1. Two images are formed.

**Question 29. A beam of parallel rays after refraction in a convex lens converges at a point.If a concave lens of the same focal length is placed lit contact with the convex lens where will the image be shifted?**

- At infinity
- At 2f
- Between 0 and f
- Between f and 2f

**Answer:** 1. At infinity

**Question 30. The ratio of powers of a thin convex and a thin concave lens is \(\frac{3}{2}\). When they are to contact, the equivalent focal length is 30 cm, Individual focal lengths are**

- 75 cm – 50 cm
- 10 cm -15 cm
- 15 cm -10 cm
- 50 cm – 75 cm

**Answer:** 2. 10 cm -15 cm

**Question 31. Two thin equal-convex lenses each of focal length 0,2 m are placed coaxially with their optical centres 0.5 in apart. Then the focal length of the combination**

- – 0.4 m
- 0.4 m
- -0.1 m
- 0.1 m

**Answer:**

**Question 32. Mercury (Hg) is coated on the plane part of a plano-convex lens. The refractive index of the lenses μ and the radius of curvature is R. The system behaves as u concave mirror whose radius of curvature is**

- μR
- \(\frac{R}{2(\mu-1)}\)
- \(\frac{R^2}{\mu}\)
- \(\frac{(\mu+1)}{(\mu-1)} n\)

**Answer:** 2. \(\frac{R^2}{\mu}\)

**Question 33. If the distance between the object and the screen is less than four times the focal length of the lens then,**

- A real image of the object will be formed on the screen for only one particular position of the lens
- Two virtual images of the object will be formed on the screen for two different positions of the lens
- No image will be formed on the screen for any position of the lens
- Two real images of the object will be formed on the screen for two different positions of the lens

**Answer:** 3

**Question 34. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describes best the image formed of an object of height 2 cm placed 30 cm from the lens?**

- Virtual, upright, height = 1 cm
- Virtual, upright, height = 0.5 cm
- Real, inverted, height = 4 cm
- Real, inverted, height =1 cm

**Answer:** 3. Real, inverted, height = 4 cm

**Question 35. A convex lens forms a real image of an object for its two different positions on the screen. If the height of the images in two cases are 8 cm and 2 cm, then the height of the object will be.**

- 2cm
- 4cm
- 8cm
- 16cm

**Answer:** 2. 4cm

**Question 36. If x _{1} is the size of the magnified image and x_{2} is the size of the diminished image in the lens displacement method, then the size of the object is**

- \(\sqrt{x_1 x_2}\)
- x
_{1}x_{2} - x²
_{1}x_{2} - x
_{1}x²_{2}

**Answer:** 1. \(\sqrt{x_1 x_2}\)

**Question 37. To determine the focal length of a thin convex lens, if red light is used instead of blue light the focal length of the lens**

- Increases
- Remains same
- Decreases
- Cannot be determined

**Answer:** 1. Increases

**Question 38. Two thin lenses of focal lengths f _{1}, and f_{2} are kept in contact co-axially. The power of the combination is given by**

- \(\sqrt{\frac{f_1}{f_2}}\)
- \(\sqrt{\frac{f_2}{f_1}}\)
- \(\frac{f_1+f_2}{2}\)
- \(\frac{f_1+f_2}{f_1 f_2}\)

**Answer:** 4. \(\frac{f_1+f_2}{f_1 f_2}\)

**Question 39. An asymmetric double convex lens is cut in two equal parts by . a plane perpendicular to the principal axis. If the power of the original lens is 4 D, the power of the halved portion will be**

- 2D
- 3D
- 4D
- 5D

**Answer:** 2D

**Question 40. A thin glass (refractive index, (μ =1.5) lens has an optical power of -5 D in air. Its optical power in a liquid medium with a refractive index of 1.6 will be**

- 1D
- -1D
- 25D
- -25D

**Answer:** 1D

**Question 41. A plano-concave lens of glass (μ = 1.5) has a radius of curvature of its curved face of 50 cm. The power of the lens is**

- -1.0 D
- – 0.2 D
- 1.0 D
- 0.2 D

**Answer:** 1 . -1.0 D

**Question 42. A plano-concave lens is made of glass of a refractive index of 1.5 and the radius of curvature of its curved surface is**

- +0.5D
- – 0.5D
- – 2D
- +2D

**Answer:** 2. – 0.5D

**Question 43. The f-number of a lens is f-15. By this statement we mean**

- Focal length of lens = \(\frac{1}{15} \times\) aperture of the lens
- Aperture of the lens = \(\frac{1}{15} \times\) Focal length of lens
- Aperture of the lens = \(\frac{1}{15^2}\) Focal length of lens
- Focal length of lens = \(\frac{1}{15^2}\) aperture of the lens

**Answer:** 2. Aperture of the lens = \(\frac{1}{15} \times\) Focal length of lens

**Question 44. Two thin lenses of focal lengths 20 cm and 25 cm are placed in contact The effective power of the combination is**

- 9D
- 2D
- 3D
- 7D

**Answer:** 1. 9D

**Question 45. The radius of curvature of the curried surface of a planoconvex lens is 20 cm. If the tire refractive index of the material of the lens is 1. 5, it will**

- Act as a convex lens only for the objects that lie on its curved side
- Act as a concave lens for objects that lie on its curved side
- Act as a convex lens irrespective of the side on which the object lies
- Act as a concave lens irrespective of the side on which the object lies

**Answer:** 3. Act as a convex lens irrespective of the side on which the object lies

**Question 46. An object approaches a convex lens from the left of the lens with a uniform speed of 5m. s ^{-1} and**

**stops at the focus. The image**

- Moves away from the lens with a uniform speed 5m. s
^{-1} - Moves away from the lens with uniform acceleration
- Moves away from the lens with a non-uniform acceleration
- Moves towards the lens with a non-uniform

**Answer:** 3. Moves away from the lens with a non-uniform acceleration

**Question 47. A convex lens Is used to form an image on a screen When the upper half of the Jens is converted by an opaque screen, then**

- Half of the Image will disappear
- A complete Image will be formed
- The intensity of the Image will decrease
- Intensity will remain the same

**Answer:** 2 And 3

**Question 48. A convex lens of the glass of refractive index μ _{1 }is immersed in a liquid of refractive index μ_{2}. It will behave as**

- Converging lens if μ
_{1}> μ_{2} - Converging lens if μ
_{1}< μ_{2} - Diverging lens if μ
_{1}> μ_{2} - Diverging Jens if μ
_{1}< μ_{2}

**Answer:** 1 And 4

**Question 49. An object and a screen are fixed at a distance d apart. When a lens of focal length f is moved between the object and the screen, sharp images of the object are formed on the screen for two positions of the lens. The magnifications produced at these two positions are M _{1} and M_{2}**

- d>2f
- d>4f
- M
_{1}M_{2} - |M
_{1}|- |M_{2}| = 1

**Answer:** 2 And 3

**Question 50. A thin, symmetric double-convex lens of power P is cut into thin parts A, B, and C as, the power of**

- A is P
- A is 2P
- B is \(\frac{P}{2}\)
- B is \(\frac{P}{4}\)

**Answer:** 1 And 3

**Question 51. Consider three converging lenses L _{1} , L_{2} and L_{3} having identical geometrical construction. The index of refraction of L_{1} and L_{2} are μ_{1} and μ_{2} respectively. The upper half of the lens L_{3} has a refractive index of μ_{1 } and the lower half has μ_{1}, A point object’ O is imaged at O_{1} by the lens L_{1} and at O_{2} by the lens placed in the same position. If L_{3} is placed in the same place**

- There will be an image at O
- There will be an image at O
_{2} - The only image will form somewhere between Oj and
- The only image will form away from O
_{2}

**Answer:** 1 And 2

**Question 52. Radii of curvature of a thin concavo-convex lens are R and 2R respectively. The refractive index of the material of the lens is μ. When the lens is placed in the air the focal length**

- Depends on the direction of the incident ray from where it comes
- Will remain the same and does not depend on which direction the incident ray comes
- Will be \(\frac{R}{\mu-1}\)
- Will be \(\frac{2 R}{\mu-1}\)

**Answer:** 2 And 4

**Question 53. A luminous object is placed at D = 100 cm away from a screen. A conjugate focus is obtained due to two positions of a convex lens of 21 cm focal length, placed in between the object and screen. The gap between the two positions of the lens (A) and (B) is d. For the positions A and B respectively linear magnifications are m _{1} and m_{2}.**

- d = 40 cm
- d = 42 cm
- \(m_1=\frac{7}{3}, m_2=\frac{3}{7}\)
- \(m_1=\frac{7}{5}, m_2=\frac{5}{7}\)

**Answer:** 1 And 3

**Question 54. The distance between a source of light and a screen is 1 m. By placing a convex lens between them an image is cast on the screen. The lens is shifted through a distance of 40 cm along the line joining the source and the screen and again an image is formed on the screen.**

**1. The focal length of the lens is**

- 40cm
- 12cm
- 21cm
- 25cm

**Answer:** 3. 21cm

**2. If the length of the images are 0.66 cm and 1.51 cm respectively the length of the object is**

- 0.998 cm
- 1.3 cm
- 0.667 cm
- 0.889 cm

**Answer:** 2. 1.3 cm

**Question 55. A convex lens of power 5 D is placed on a plane mirror. A pin is placed above 30 cm straight from the lens.**

**1. The distance of the image from the lens at which the image is formed is**

- 30 cm
- 40 cm
- 50 cm
- 60 cm

**Answer:** 4. 60 cm

**2. The distance of the pin from the lens so that its image will coincide with the pin is**

- 20 cm
- 30 cm
- 40 cm
- 50 cm

**Answer:** 1. 20 cm

**Question 56. Two convex lenses of focal lengths 3cm and 4 cm are placed 8 cm apart from each other. An object of height 1 cm is placed at a distance of 4 cm from the lens’s smaller focal length**.

**1. The position of the final image formed by the two lenses**

- 8 cm away from the right side of the second lens
- 2 cm away from the left side of the second lens
- 2 cm away from the right side of the second lens
- 1 cm away from the left side of the second lens

**Answer:** 3. 2 cm away from the right side of the second lens

**2. The size of the image will be**

- 3 cm
- 2 cm
- 2.5 cm
- 18 cm

**Answer:** 4. 18 cm

**Question 57. A convex lens forms a five-times magnified real image of an object. When the object is further shifted by 6 cm away from the lens the magnification of the real image becomes double**

**1. The focal length of the lens is**

- 20 cm
- 15 cm
- 25 cm
- 18 cm

**Answer:** 1. 20 cm

**2. The distance of the object from the lens in the first case**

- 18 cm
- 24 cm
- 20 cm
- 32 cm

**Answer:** 2. 24 cm

**Question 58. The refractive index of the material of a double equi convex lens is 1****.5. If the radius of curvature of the lens is R, then its focal length is**

- Zero
- Infinite
- 2R
- R

**Answer:** 4. R

If R is the radius of curvature of one surface of the lens, the radius of curvature of the other surface =-R

Now

⇒ \(\frac{1}{f}=(\mu-1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

⇒ \((1.5-1)\left(\frac{1}{R}-\frac{1}{-R}\right)\)

⇒ \(0.5 \times \frac{2}{R}=\frac{1}{R}\)

f = R

**Question 59. A luminous object is separated from a screen by distance d is A convex lens is placed between the object and the object and the screen such that it forms a distinct image on the screen. The maximum possible focal length of this convex lens is.**

- 4d
- 2d
- \(\frac{d}{2}\)
- \(\frac{d}{2}\)

**Question 60. An object is placed 30 cm away from a convex lens of focal length 10 cm and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by 45 to get a sharp image again. The magnitude of the focal length of the concave lens is (in cm).**

- 72
- 60
- 36
- 20

**Answer:** 4. 20

Here, u_{1} = -30cm, f_{1 }= 10 cm, v_{1} = ?

For the first case,

⇒ \(\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}\)

⇒ \(\frac{1}{v_1}+\frac{1}{30}=\frac{1}{10}\)

Or, \(\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\)

⇒ \(\frac{1}{f}=\frac{1}{60}+\frac{1}{30}=\frac{1}{20}\)

Where , u_{2}= -30 cm, v_{2}= 15 + 45= 60 cm

∴ \(\frac{1}{f_2}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}\)

f_{2} = 20 cm

**Question 61. An object is located 4 m from the first of two thin converging lenses of focal lengths 2 m and lm respectively. The lenses are separated by 3 m. The final image formed by the second lens is located from the source at a distance of**

- 8.0 m
- 7.5 m
- 6.0 m
- 6.5 m

**Answer:** 2. 7.5 m

For the first convex lens, the object is placed at a distance of 2/(2 x 2 m = 4 m). The image for the first lens would be formed at a distance of 4 m in the opposite direction.

But the second lens is at a distance of 3 m, consequently, the image of the first lens would be at a distance of m behind the second lens.

Hence, the effective object distance for the second lens is u = +1m.

The focal length of the second lens, f= + 1 m

If the image distance is v,

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

Or, \(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}\)

Or, v = \(\frac{u f}{u+f}=\frac{1 \times 1}{1+1}\)

= 0. 5 m

∴ Distance of the final image from the source

= 4 + 3 + 0.5 = 7.5 m

**Question 62. A point object is held above a thin biconvex lens at its focus. The focal length is 0.1m and the lens rests on a horizontal thin plane mirror. The final image will be formed at**

- Infinite distance above the lens
- 0.1m above the centre of the lens
- Infinite distance below the lens
- 0. 1 m below the centre of the lens

**Answer:** 2. 0.1m above the centre of the lens

Here a point object is placed at focus (F) of the convex lens. Thus, a beam of diverging rays emerges from it and becomes parallel after refraction from the lens and is incidentally on the plain mirror. As a result, reflected travels the same path and again refraction from the lens meets at the focus of the lens.

**Question 63. Two identical biconvex lenses, each of focal length/ are placed side by side by side in contact with each other with a layer of water in between them as shown in the figure. If the refractive index of the material of the lenses is greater than that of water, howthecombinedfocallength F related to f**

- F>f
- \(\frac{f}{2}<F<f\)
- \(F<\frac{f}{2}\)
- F = f

**Answer:** 2. \(\frac{f}{2}<F<f\)

If the focal length of the lens formed by water be f_{w} then,

⇒ \(\frac{1}{f_w}=\left(\mu_w-1\right)\left(-\frac{1}{R}-\frac{1}{R}\right)\)

⇒ \(\left(\mu_w-1\right)\left(-\frac{2}{R}\right)\)

If the focal length of each lens is f then,

⇒ \(\frac{1}{f_l}=\left(\mu_l-1\right)\left\{\frac{1}{R}-\left(-\frac{1}{R}\right)\right\}=\left(\mu_l-1\right) \frac{2}{R}\) …………… (1)

⇒ \(\frac{1}{f_{\mathrm{eq}}}=\frac{1}{f_w}+\frac{1}{f_l}+\frac{1}{f_l}\)

= \(\left(\mu_w-1\right)\left(-\frac{2}{R}\right)+\frac{2}{f_l}\)

= \(-\frac{\mu_w-1}{\mu_l-1}\left(\frac{1}{f_l}\right)+\frac{2}{f_l}\) Using equation 1

= \(\frac{1}{f_l}\left\{2-\frac{\mu_w-1}{\mu_l-1}\right\}\) ………….. (2)

Since \(\frac{\mu_w-1}{\mu_l-1}<1\)

From eq. (2) we get \(\frac{1}{f_{\text {eq }}}<\frac{2}{f_l}\)

Therefore, \(\frac{1}{f_l}<\frac{1}{f_{\text {eq }}}<\frac{2}{f_l}\)

Or, \(f_l>f_{\text {eq }}>\frac{f_l}{2}\)

Given f_{l} = f and f_{eq} = F]

Or, \(\frac{f}{2}<F<f\)

**Question 64. A diverging lens with a magnitude of focal length 25cm is I placed at a distance of 15cm from a converging lens of a magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final image formed is 1 =**

- Real and at a distance of40cm from the convergent lens
- Virtual and at a distance of40cm from the convergent lens
- Reak and at a distance of 40cm from the convergent lens
- Real and at a distance of cm from the convergent lens

**Answer:** 1. Real and at a distance of40cm from the convergent lens

For converging lens, u = -(15 + 25) = -40cm

Since u = 2f_{2}[f_{2} = 20 cm ]

Therefore, v = u = 40crn

So, the final Image Is real and Is formed at a distance of 40cm from the converging lens,

**Question 65. Two Identical thin plano-convex glass lenses (refractive index 1.5) each having a radius of curvature of 20 cm are placed with their convex surfaces In contact at the center. The Interveningspace Is filled with oil of refractive Index1,7. The focal length of the combination is**

- – 20 cm
- – 25 cm
- – 50 cm
- 50 cm

**Answer:** 3. – 50 cm

We know \(\frac{1}{f}=(\mu-1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

For the lens on the left,

⇒ \(\frac{1}{f_1}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{-20}\right)=\frac{1}{40}\)

For the lens on the right,

⇒ \(\frac{1}{f_2}=(1.5-1)\left(\frac{1}{+20}-\frac{1}{\infty}\right)=\frac{1}{40}\)

For the oil in the Intervening space

⇒ \(\frac{1}{f_3}=(1.7-1)\left(\frac{1}{-20}-\frac{1}{20}\right)=-\frac{7}{100}\)

Hence, if the focal length of the combination is F, then

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{1}{40}+\frac{1}{40}-\frac{7}{100}=-\frac{1}{50}\)

I. e F = – 50 cm

## Refraction Of Light At Spherical Surface Lens Synopsis

1. Usually the transparent refracting medium is bounded by two spherical surfaces or a spherical surface and a plane surface. is. called a **lens**

- If the refractive index for the material of a lens is greater than the refractive index of its surrounding medium, then a convex lens behaves as a converging lens and a concave lens behaves as a diverging lens
- If the refractive index for the material of a lens is less than the refractive index of its surrounding medium, then a convex lens behaves as a diverging lens and a concave lens behaves as a converging lens

2. If the thickness of a lens is much less than the radii of curvature of two surfaces then that lens is called a thin lens

3. The point situated on the principal axis of the lens and within the lens through which rays oflight emerge without deviation after refraction, is called the optical centre of the

lens

4. **A lens has two foci:**

- The first principal focus,
- Second principal focus. Usually, the principal focus of a lens means the second principal focus.

5. The distance between the principal focus and the optical centre of a lens is called the focal length (f) of the lens

6. The ratio of the length of the image to that of the object is called linear magnification of the image.

7. The ratio of the area of the image to that of the two-dimensional object is called areal magnification of the image

8. The ratio of the length of the image to that of the object along the principal axis is called longitudinal or axial magnification.

9. A convex lens always forms a real image for a virtual object and that image always lies within the focus.

10. If the object distance for a virtual object in front of a concave lens be less than its focal length then the image is always real.

11. A pair of points is such that if an object is placed at one of them, its image is formed at the other by a fixed lens, then that pair of points are called conjugate foci of that lens

12. If the surrounding medium of a lens is denser than air then the focal length of a lens increases.

13. An image of an object is formed by a combination of a number of co-axial lenses. Now, keeping the position of the object and the image unaltered a single lens is used instead of a combination.

14. If the magnification of the image thus produced by the lens is equal to that in the case of the combination of lenses then this single lens is called the equivalent lens of the combination of lenses.

15. The degree of convergence or divergence of parallel rays of light incident on a lens gives the measure of the power of the lens.

16. The general form of refraction of light at the common spherical surface, when light rays moving from a medium of refractive index fix are refracted in the medium of refractive index μ_{2}, is

⇒ \(\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}\)

Where, u = object distance; V = image distance and R = radius of curvature of spherical surface

17. General equation of a lens \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

This is also called the conjugate foci relation

18. Linear magnification

m = \(\frac{\text { length or height of the image }}{\text { length or height of the object }}\)

= \(\frac{v}{u}=\frac{f-v}{f}=\frac{f}{u+f}\)

19. Areal magnification

m’ = \(\frac{\text { area of the image }\left(A^{\prime}\right)}{\text { area of the object }(A)}=m^2\)

20. Longitudinal or axial magnification

m” = \(\frac{\text { length of the image along the principal axis }}{\text { length of the object along the principal axis }}\)

= \(\frac{d v}{d u}=-m^2\)

= – m²

21. Newton’s equation, xy = f²

Where x = u- f and y = v- f

22. Lens maker’s formula \(\frac{1}{f}=\left({ }_a \mu_b-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)

Where, _{a}μ_{b} = refractive index of the material of the lens concerning the surrounding medium a, rx = radius of curvature of the first refracting surface of the lens, r_{2 }radius of curvature of the second refracting surface of the lens.

23. Two thin coaxial lenses have focal lengths f_{1} and f_{2} respectively, the distance between their optical centres if the equivalent focal length of this combination of lenses is F then

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{a}{f_1 f_2}\)

If a = 0, i.e., when the lenses are in contact then

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\)

24. A convex lens is placed in between an object and a screen which are kept at a fixed distance D. In this case

If D > 4f, then for two different positions of the lens two real images will be formed on the screen.

If D = 4f, then for a fixed position of the lens a real image will be formed on the screen.

If D < 4f, then no image will be formed on the screen for any position of the lens.

If D > 4f, then f= \(\frac{D^2-x^2}{4 D}\) where x = distance between the two positions of the lens

If D > 4f, then u_{1} = v_{2} and v_{1}= u_{2}, where u_{1} and v_{1 }are the object and image distance respectively for the first position of the lens and u_{2} and v_{2} are the object and image distance respectively for the second position of the lens.

1f D > 4f, then d = \(\sqrt{d_1 d_2}\), where d- the size of the object, dy and d2 = size of the two images for two different positions of the lens

Power of a lens, P = + \(+\frac{100}{f}\) dioptre {wheref is in cm]

= + \(\frac{1}{f}\) dioptre {wheref is in m]

If two lenses have power P_{1} and P_{2} then when these two lenses are kept in contact with each other, the power of their combination, P = P_{1}+P_{2}