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

If the magnitude of dispersive power of two lenses are $0.024$ and $0.036$. There focal length will be for abberation free combination.

  1. $30\ cm,\ -40\ cm$
  2. $30\ cm,\ -45\ cm$
  3. $10\ cm,\ 30\ cm$
  4. $20\ cm,\ -35\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an aberration-free combination of two thin lenses, the condition is w1/f1 + w2/f2 = 0, where w is the dispersive power and f is the focal length. Given w1=0.024 and w2=0.036, the ratio f1/f2 = -w1/w2 = -0.024/0.036 = -2/3. Option B provides f1=30 cm and f2=-45 cm, which satisfies 30/-45 = -2/3.

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

In displacement method,magnification for two positions of the lens are 2 and 0.5 and the distance between the two position of the lens is 30 cm. if the focal length of lens is

  1. 15cm

  2. 20cm

  3. 25cm

  4. 30cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the displacement method, the focal length f is given by f = d(1-m^2) / (m1-m2) is incorrect; the correct formula is f = D*m / (1+m)^2 or using the displacement d and magnifications m1=2, m2=0.5. Since m1*m2 = 1, the object distance u and image distance v are swapped. The distance between positions is d = v-u = 30. With m=v/u=2, v=2u. Then 2u-u=30, so u=30, v=60. Focal length f = (u*v)/(u+v) = (30*60)/(90) = 20 cm.

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

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
A Correct answer
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.