A magnification of 2 indicates the image is twice the size of the object and a magnification of 1 indicates an image size being the same as the object size. The sign conventions for the given quantities in the lens equation and magnification equations are as follows: f is + if the lens is a double convex lens (converging lens) f is - if the lens is a double concave lens (diverging lens) d i is + if the image is a real image and located on the opposite side of the lens. On the other hand, the magnification m is negative when the image formed is real and inverted. An object 5 cm high is held 25 cm away from a converging lens of focal length 20 cm. u   =   object distance Please note that the magnification formula is applicable both in convex lenses and concave lenses. Code to add this calci to your website . On the other hand, the magnification m is negative when the image formed is real and inverted. Calculate the position of the images formed by the fol-lowing concave lenses. In the case of a concave lens, it is always positive. Magnification of Convex Lens: It is a ratio between the image height and object height. (click on the green letters for the solutions). If the magnification is positive, then the image is upright compared to the object (virtual image). An expression showing the relation between object distance, image distance and focal length of a mirror is called mirror formula. Example Problem #1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Learn more about Reflection of Light here. Section 3: Concave Lenses 14 Here are some exercises with concave lenses. Q. Please note that the magnification formula is applicable both in convex lenses and concave lenses. Using lens formula the equation for magnification can also be obtained as . Therefore, the relationship between the object distance, the image distance and the focal length of a lens is given by the $$\text{Lens Formula: }\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$ The lens formula may be applied to convex lenses as well as concave lenses provided the ‘real is positive’ sign convention is followed. (a)For an object with u= 12cmif the focal length is 4cm Solution: As the lens here is concave, the focal length (f) = -10 cm. Hence, the expression for magnification (m) becomes: m = h’/h = -v/u. The focal length of the mirror is f. a. Determine the image distance and the image size. Be careful with the sign of the focal length in the lens formula! $$\frac{1}{m(m+1)}$$ b. Formula : Where, f - Focal length, d i - Image distance, d 0 - Object distance. Is lens formula applicable only for convex lens? Assumptions and Sign conventions Draw the ray diagram and find the position, size and nature of the image formed. If the magnification is positive, then the image is upright compared to the object (virtual image). The magnification ($$M$$) of the image formed can be calculated using the following formula. Example 2: The distance of an object of height 6 cm from a concave lens is 20 cm. The two formulas given above are together referred to as the thin lens formula. We also have another formula for magnification in lenses Magnification = v/u where v is image distance u is object distance Note: - If magnification (m) is positive, It means image formed is virtual and erect If magnification (m) is negative, It means image formed is real and inverted Questions Example 10.3 - A concave lens has focal length of 15 cm. As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution. Physics Grade XI Reference Note: Mirror formula for concave mirror when real image is formed and for convex mirror. What will be the distance of the object, when a concave mirror produces an image of magnification m? Online physics calculator that calculates the concave mirror equation from the given values of object distance (do), the image distance (di), and the focal length (f). If ‘ $$i$$ ’ is positive, the image is upright and if ‘ $$i$$ ’ is negative, the image is inverted. Exercise 2. Magnification of Convex Lens: It is a ratio between the image height and object height. Object distance (o) = + 20 cm Now by putting the values of (f) and (o) in 1/v+ 1/o = 1/f, we will get, 1/v = 1/f - 1/o Where, $$v$$ is the object height $$u$$ is the image height. At what distance should the object from the lens be placed so … And the magnification m is positive when the image formed is virtual and erect. A magnification of 2 indicates the image is twice the size of the object and a magnification of 1 indicates an image size being the same as the object size. Lens formula is applicable for convex as well as concave lenses. In the case of a concave lens, it is always positive. It is an equation that relates the focal length, image distance, and object distance for a spherical mirror. If its focal length is 10 cm, calculate the size and position of the image formed. Solved Example for You. m = h 2 /h 1 = v//u = (f-v)/f = f/(f+u) This equation is valid for both convex and concave lenses and for real and virtual images. You can try the following sample problem using this. It is given as, \frac {1} {i} + \frac {1} {o} = \frac {1} {f} i= distance of the image from the lens. Define lens formula. A concave lens of focal length 15 cm forms an image 10 cm from the lens. If magnification is negative then the image is inverted as … These lenses have negligible thickness. Using lens formula the equation for magnification can also be obtained as m = h2/h1 = v//u = (f-v)/f = f/ (f+u) This equation is valid for both convex and concave lenses and for real and virtual images. Table shows the sign convention for the values of object distance, image … And the magnification m is positive when the image formed is virtual and erect. The magnification is negative for real image and positive for virtual image.