Examkin

Soru

Zorluk: Çok zorReflection of Light at Plane and Curved Mirrors

A convex mirror forms an upright image that is 13\frac{1}{3} the size of an object. When the object is moved 20 cm20\text{ cm} further away from the mirror, the size of the image becomes 15\frac{1}{5} the size of the object. What is the radius of curvature of the mirror?

  1. 20 cm20\text{ cm}Cevap
  2. B
    10 cm10\text{ cm}
  3. C
    40 cm40\text{ cm}
  4. D
    5 cm5\text{ cm}

Cevap

The radius of curvature of the convex mirror is 20 cm20\text{ cm}.
For a convex mirror, the focal length is negative (f=−f0f = -f_0). The magnification formula m=ff−um = \frac{f}{f - u} for a virtual upright image gives m=−f0−f0−u=f0f0+um = \frac{-f_0}{-f_0 - u} = \frac{f_0}{f_0 + u}. For m1=13m_1 = \frac{1}{3}, we get u1=2f0u_1 = 2f_0. For m2=15m_2 = \frac{1}{5}, we get u2=4f0u_2 = 4f_0. The object displacement is u2−u1=2f0=20 cmu_2 - u_1 = 2f_0 = 20\text{ cm}, which gives f0=10 cmf_0 = 10\text{ cm}. The radius of curvature is R=2f0=20 cmR = 2f_0 = 20\text{ cm}, making the choice stating 20 cm20\text{ cm} correct.

Adım Adım Çözüm

1
Apply the magnification formula and sign convention for a convex mirror at the first position.
For a convex mirror, f=−f0f = -f_0 and the image is virtual (v1=−v01v_1 = -v_{01}). Magnification m1=−v1u1=v01u1=13m_1 = -\frac{v_1}{u_1} = \frac{v_{01}}{u_1} = \frac{1}{3}, so v01=u13v_{01} = \frac{u_1}{3}.
Convex mirrors always form virtual, upright, and diminished images.
2
Substitute v1v_1 into the mirror equation for the first position to express u1u_1 in terms of focal length magnitude f0f_0.
1−f0=1u1+1−v01=1u1−3u1=−2u1  ⟹  u1=2f0\frac{1}{-f_0} = \frac{1}{u_1} + \frac{1}{-v_{01}} = \frac{1}{u_1} - \frac{3}{u_1} = -\frac{2}{u_1} \implies u_1 = 2f_0.
The mirror formula is 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} with cartesian sign conventions.
3
Repeat the mirror equation calculation for the second object position.
For m2=15m_2 = \frac{1}{5}, v02=u25v_{02} = \frac{u_2}{5}. Substituting gives 1−f0=1u2−5u2=−4u2  ⟹  u2=4f0\frac{1}{-f_0} = \frac{1}{u_2} - \frac{5}{u_2} = -\frac{4}{u_2} \implies u_2 = 4f_0.
The second object position gives a magnification of 15\frac{1}{5}.
4
Use the known object shift distance to solve for f0f_0 and radius of curvature RR.
u2−u1=20 cm  ⟹  4f0−2f0=20 cm  ⟹  2f0=20 cm  ⟹  f0=10 cmu_2 - u_1 = 20\text{ cm} \implies 4f_0 - 2f_0 = 20\text{ cm} \implies 2f_0 = 20\text{ cm} \implies f_0 = 10\text{ cm}. Since R=2f0R = 2f_0, R=20 cmR = 20\text{ cm}.
The distance between the two object positions is 20 cm20\text{ cm}, and the radius of curvature of a spherical mirror is twice its focal length.

Anahtar Kavram

Spherical Mirror Formula and Sign Conventions for Convex Mirrors
Bu soruyu puanla

Konular

Reflection of Light at Plane and Curved Mirrors

Bu konudan daha fazla soru