Tag: ray optics and optical instruments

Quizzes (21) Questions (129)

Questions Related to ray optics and optical instruments

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

The refractive index of the material of a double convex lens is $1.5$ and its focal lengths in $5cm$. If the radii of curvature are equal, the value of the radius of curvature is

  1. 5.0

  2. 6.5

  3. 8.0

  4. 9.5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the lens maker formula 1/f = (n-1)(1/R1 - 1/R2). For a double convex lens, R1=R and R2=-R. So 1/5 = (1.5-1)(1/R + 1/R) = 0.5 * (2/R) = 1/R. Thus R = 5 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

In a plano convex lens, the radius of curvature of the convex iens is 10 cm, if the plane side is polished , then the focal length is (Refractive index=1.5)

  1. $20.5 cm$
  2. $10 cm$
  3. $15.5 cm$
  4. $5 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a plano-convex lens with the plane side polished, it behaves like a combination of a lens and a mirror. The equivalent focal length is given by 1/f = 2/f_lens. With R = 10 cm and n = 1.5, f_lens = R/(n-1) = 20 cm, making f = 10 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

When a thin convex lens of focal length $10cm$ is kept in contact with a diverging lens, the power of the combination is found to be $-10D$. The focal length of the other lens is

  1. $-5cm$
  2. $10cm$
  3. $-25cm$
  4. $5cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power P = P1 + P2. The power of the first convex lens is 1/0.1 = +10 D. The total power is given as -10 D. Thus, -10 = 10 + P2, giving P2 = -20 D. The focal length of the second lens is f2 = 1/P2 = 1/(-20) m = -5 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

A lens made from a material of absolute refractive index $\mathrm { n } _ { 1 }$  and it is placed in a medium of absolute refractive index $\mathrm { n } _ { 2 }$   The focal length of the lens is related to $\mathrm { n } _ { 1 } \text { and } \mathrm { n } _ { 2 }$ as:

  1. $f \alpha \left( n _ { 1 } - n _ { 2 } \right)$
  2. $f \alpha \frac { 1 } { \left( n _ { 1 } - n _ { 2 } \right) }$
  3. $f \alpha \left( n _ { 1 } + n _ { 2 } \right)$
  4. $f \alpha \frac { 1 } { \left( n _ { 1 } + n _ { 2 } \right) }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to the lens maker's formula, 1/f = ((n1/n2) - 1) * (1/R1 - 1/R2). Simplifying this term shows that focal length f is inversely proportional to (n1 - n2).

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Two thin convex lenses of focal length 10 cm  and 15 cm are separated by a distance of 10 cm. The  focal length of combination is   :-

  1. 4.2 cm

  2. 6 cm

  3. 10 cm

  4. 15 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The effective focal length of two lenses separated by a distance d is given by 1/F = 1/f1 + 1/f2 - d/(f1*f2). Substituting f1 = 10 cm, f2 = 15 cm, and d = 10 cm gives 1/F = 1/10 + 1/15 - 10/(10*15) = 3/30 + 2/30 - 2/30 = 3/30 = 1/10, so F = 10 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

A point object is placed at a distance of $15 cm$ from a convex lens. The image is formed on the other side at a distance of $30cm$ from the lens. When a concave lens is placed in contact with the convex lens, the image shifts away further by $30 cm$. Calculate the focal lengths of the concave and convex lenses.

  1. $10 cm, 60 cm$
  2. $ 20 cm, 30 cm$
  3. $60 cm, 10 cm$
  4. $30 cm, 20 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the convex lens, 1/v - 1/u = 1/f. 1/30 - 1/-15 = 1/f1 => 1/30 + 2/30 = 3/30 = 1/10, so f1 = 10 cm. With the concave lens, the image shifts by 30 cm, so the new image distance is 60 cm. 1/60 - 1/-15 = 1/F_eq => 1/60 + 4/60 = 5/60 = 1/12, so F_eq = 12 cm. Since 1/F_eq = 1/f1 + 1/f2, 1/12 = 1/10 + 1/f2 => 1/f2 = 1/12 - 1/10 = (5-6)/60 = -1/60. So f2 = -60 cm. The focal lengths are 10 cm and -60 cm.

Multiple choice physics ray optics and optical instruments power of the lens thin lens combination of lenses

Two plano-convex lenses of glass of refractive index $1.5$ have radii of curvature $20\ cm$ and $30\ cm$. They are placed in contact with curved surfaces towards each other and the space between them is filled with a liquid of refractive index $4/3$. The focal length of the combination is

  1. $48\ cm$
  2. $72\ cm$
  3. $12\ cm$
  4. $28\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The system consists of two plano-convex lenses and a liquid lens. P_total = P1 + P2 + P_liquid. P1 = (1.5-1)/0.2 = 2.5D. P2 = (1.5-1)/0.3 = 1.67D. P_liquid = -2*(n_liq-1)/R_avg? No, the liquid lens is biconcave with radii 20 and 30. P_liq = (4/3 - 1) * (-1/20 - 1/30) = (1/3) * (-5/60) = -5/180 = -1/36. P_total = 1/40 + 1/60 - 1/36 = (9+6-10)/360 = 5/360 = 1/72. So f = 72 cm.