JPH0216492B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a single-lens reflex finder, and more specifically, the present invention relates to a single-lens reflex finder, and more particularly, a mirror for bending the optical axis of light for the view finder is configured so that its reflective surface passes through the second principal point of the view finder master lens. This invention relates to a single-lens reflex viewfinder that is extremely easy to adjust by installing it.

As is well known, it is desirable that the imaging side screen of the camera and the viewfinder screen match, but in a twin-lens reflex camera, etc., the light entrance for imaging and the light entrance for the viewfinder are the same. In the case of cameras with different types, it is theoretically impossible to make both screens completely coincident.

Single-lens reflex viewfinders, which separate a portion of the light focused by the camera's main lens and guide it to the viewfinder, are able to align both screens perfectly, and many modern cameras use this method. I am using it. In particular, cameras for videotape recorders (VTR) have a smaller image size than 35mm cameras, so misalignment between the two screens is more likely to occur.Moreover, images are often taken while using a monitor television, which causes misalignment between the two screens. Single-lens reflex viewfinders are often used because they can be seen immediately.

Figure 1 shows the results using a single-lens reflex viewfinder.
1 is a diagram showing a schematic structure of a VTR camera.
In the figure, the front part of the camera 1 (the right side in the figure) has a zoom lens 2, which is the main lens of the camera.
is attached, and the light focused by the ahocal system of the zoom lens 2 is directed along the optical axis (hereinafter this optical axis will be referred to as the "zoom optical axis") A ₁ -
Proceed along A ₂ . A prism 3 is installed on the zoom optical axis _A1 - _A2 , and one surface 3a of the prism 3 forms an angle of 45° with respect to the zoom optical axis _A1 - _A2. Because half mirror processing is applied, part of the light focused by the zoom lens 2 continues along the zoom optical axis _A1 - _A2 , but part of it is reflected and its path is bent by 90 degrees. is,
Optical axis B ₁ - _{B 2} (hereinafter referred to as "VF master optical axis")
proceed along. A viewfinder master lens 4 is installed on the VF master optical axis B ₁ - B ₂ , and beyond that, a viewfinder master lens 4 is installed on the VF master optical axis B ₁ - _{B 2.}
A mirror 5 having a reflective surface forming an angle of 45 degrees is installed, and the action of this mirror 5 bends the course of the light by 90 degrees again, forming an optical axis C ₁ -C ₂ (hereinafter referred to as the "VF optical axis").
That's what it means. ). On the VF optical axis C ₁ - C ₂ there is a view finder first imaging point 6 formed by the view finder master lens 4, and beyond that on the VF optical axis C ₁ - C ₂ there is a field lens 7,
A relay lens 8 is installed, whereby a viewfinder second imaging surface 9 is present, on which a VF mask 10 is provided. Further, on the VF optical axis C ₁ - C ₂ , there is an eyepiece 11a.
An eyepiece section 11 is provided.

A master lens 21, a crystal filter 22, and a vidicon 23 are installed along the zoom optical axis _A1 - _A2 from the prism 3, and the end surface of the vidicon 23 close to the zoom lens 2 (the right side in the figure) is the imaging plane. 23a.

In the case of the camera using the single-lens reflex viewfinder described above, the light forming the viewfinder screen and the light forming the camera's imaging surface screen are both collected by the zoom lens 2.
The two screens should theoretically match perfectly.

However, the above explanation assumes that each part used in this single-lens reflex viewfinder is manufactured and assembled accurately as designed, and there are no manufacturing errors or errors.
When there is an assembly difference, the two screens do not necessarily match. For example, in Figure 2, the prism is designed to form an angle of 45° with respect to the zoom optical axis A ₁ - _{A 2} (therefore, it is also 45° with respect to the VF master optical axis B ₁ - _{B 2} ). This figure shows the case where one surface 3a of 3 forms an angle θ (θ≠45°) with respect to the VF master optical axis B ₁ −B ₂ , and the light beam 31 focused by the zoom lens 2 is A light beam 31 reflected by the surface 3a and tilted at a predetermined angle with respect to the VF master optical axis B ₁ -B ₂
b and viewfinder master lens 4
into the viewfinder master lens 4
The luminous flux once travels parallel to the VF master optical axis B ₁ -B ₂ between the first principal surface 4a and the second principal surface 4b, but then proceeds parallel to the incident angle again.
After passing through the viewfinder master lens 4, it is reflected by a mirror 5 and is imaged at a point Q on the viewfinder first imaging plane 6. On the other hand, when the surface 3a is installed accurately as designed, the light beam becomes a light beam 31a, which similarly forms an image at the point _Q0 . This phenomenon in which an image is formed with a deviation of Δ= ₀ from the normal image is called optical axis misalignment. Also mirror 5 and prism 3
is installed at an angle to the design value,
The screen viewed from the Viewfinder appears to be tilted towards the subject. This phenomenon is called screen tilt.

It is desirable to eliminate these optical axis deviations and screen inclinations, so adjustments must be made. Furthermore, even if there is no prism assembly error as described above, there may be cases where it is necessary to intentionally adjust the optical axis to cause an optical axis shift. For example, even if the center of the monitor (the center of the image plane 23a) is aligned with the zoom optical axis A ₁ -A ₂ , the center of the VF mask 10 that forms the viewfinder screen is aligned with the VF optical axis C ₁ -C _{2 .}
If it is misaligned, the center of the TV monitor and the Viewfinder screen do not match, and you will need to make adjustments.

When making this adjustment in a conventional single-lens reflex viewfinder, a method as shown in FIG. In the figure, a mirror holder 12 is attached to a bracket 1a attached to a camera 1 with screws 14 via a spring 13. Also, bracket 1a has 3
Adjustment screws 15 are screwed together, and the tip of each adjustment screw 15 is in contact with one surface of the mirror holder 12. A mirror 5 is attached to the other surface of the mirror holder 12. By adjusting this adjustment screw 15, the inclination of the mirror 5 can be changed, and optical axis deviation, screen inclination, etc. can be adjusted.

However, in the conventional adjustment method described above, when attempting to correct the optical axis deviation in the horizontal direction, the optical axis in the vertical direction also moves, so adjustments must be made by repeating the trial and error method several times. I had to. Conventionally, the inclination of not only the mirror 5 but also the prism 3 is changed in a manner similar to that shown in FIG. 3, and a method has been adopted in which adjustment is performed using the mirror 5 and the prism 3. However, even with this method, if the optical axis shift is corrected in a situation where the screen tilt is 0, the screen tilt will occur, and conversely, if the screen tilt is corrected in the situation where the optical axis shift is 0, the optical axis shift will occur. It was necessary to repeat the process of trial and error, and in any case, it had the disadvantage of requiring a lot of manual labor to adjust it to within the standard.

The present invention eliminates the drawbacks of the conventional single-lens reflex viewfinder adjustment method described above, and has found an adjustment method for a single-lens reflex viewfinder that can easily adjust optical axis deviation, screen tilt, etc., and furthermore, the discovered adjustment method. The purpose of the present invention is to provide a single-lens reflex viewfinder that makes it easier to make adjustments based on .

The present invention deals with optical axis deviation of a single-lens reflex viewfinder,
Focusing on the fact that screen tilt etc. are mainly caused by poor mounting angles of the reflective surfaces of mirrors and prisms, we optically analyzed the behavior of light in that case and determined the focal length fm of the viewfinder master lens,
We discovered that when the distance between the intersection of the mirror and the optical axis and the first image forming surface of the viewfinder is the same, the optical axis deviation becomes zero by itself by adjusting only the screen tilt. The above object is achieved by arranging the reflective surface of the lens so as to pass through the second principal point of the viewfinder master lens.

Embodiments of the present invention will be described below based on the drawings. First, when considering how to adjust a single-lens reflex viewfinder, it is important to remember that optical axis misalignment, screen tilt, etc. are mainly caused by improper installation angles of the mirror and prism, so all parts other than the prism and mirror are manufactured as designed. I will proceed with the argument assuming that it has been assembled accurately. Further, it is assumed that the prism has no aberration. Furthermore, if the prism 3 is tilted in the path of the light ray shown in FIG. 2, there will be some refraction when entering and exiting the prism, but since the focal point will not change, the refraction is ignored when drawing.
Although the entire prism is not tilted, if only the half-mirror surface of the prism is slightly tilted, the imaging point will shift slightly due to refraction when the prism enters and exits the prism. Since this movement is extremely small compared to the effect of optical axis deviation due to the inclination of the surface of the half mirror process, we will proceed with the argument assuming that there is no error.

First, let us consider the case where the mirror is provided at a normal position and only the prism is tilted. That is, in FIG. 4, the zoom optical axis A ₁ −
The point on the ray 41 that coincides with A ₂ is Q ₁ , and the virtual image of Q ₁ by the half mirror surface 3a of the prism 3 is Q ₂
shall be. Ray 41 passes through Q ₁ and zooms optical axis A ₁ −
A part of the light is reflected toward the viewfinder master lens ₄ by the half mirror surface 3a at the intersection O1 between A2 and the VF master _optical axis _B1 - _B2 . The direction of this reflection is the direction of the line connecting Q ₂ O ₁ . Yotsute ray 41
The inclination δ between the VF master optical axis B ₁ -B ₂ is expressed as a function of the inclination (θ-45) of the half mirror surface 3a from the reference angle of 45° and the point Q ₁ . Here θ
is VF master - optical axis B ₁ - B ₂ and half mirror surface 3
It is the angle formed by a.

Consider a ray 42 that is parallel to this ray 41 and whose extension of the incident light to the viewfinder master lens 4 passes through the first principal point _K1 of the viewfinder master lens 4. The imaging point of this light ray 42 and the light ray 41
Since the imaging points of the rays 42 and 42 are the same, finding the imaging point of the ray 42 means finding the imaging point of the ray 41. The light ray 42 emerges from the viewfinder master lens in a straight line forming an angle of δ from the second principal point K ₂ of the viewfinder master lens 4 . This light beam is reflected by the mirror 5 toward the eyepiece 11a. Mirror 5 of point K ₂
If the virtual image is K ₂ ', then the direction of reflection of the light ray 42 by the mirror passes through K ₂ ' and is in the direction of a straight line forming an angle δ with respect to the VF optical axis C ₁ -C ₂ . Focal length of viewfinder master lens 4 from K ₂ '
It intersects with the viewfinder first imaging point 6 at a point extending along the VF optical axis C ₁ -C ₂ by fm toward the eyepiece section. This point Q becomes the imaging point of a group of parallel rays including the rays 41 and 42. This allows the optical axis deviation Δ to be determined.

The above is a two-dimensional analysis, but the same applies to a three-dimensional case. The formula for determining the optical axis deviation will be derived below based on the above-mentioned concept. As shown in Fig. 5, the coordinate axis x is on the optical axis _A1 - _A2 , the coordinate axis z is on the VF master optical axis _B1 - _B2 , and the axis perpendicular to x and z is y, O1 _. Consider the coordinates (x, y, z) with the origin as the origin. This xyz coordinate is rotated by the angle Ψ around the x axis, and then the rotated y ₂
The coordinates generated by rotating by the angle θ around the axis are set as coordinates (x ₂ , y ₂ , z ₂ ), and the half mirror processing surface 3a
Let be a plane that includes the y ₂ and z ₂ axes.

Now, any point Q ₁ in space is expressed as xyz coordinates (x,
y, z), and let the x ₂ y ₂ z ₂ coordinates of this point Q ₁ be (x ₂ , y ₂ , z ₂ ). At this time,
The coordinates of the virtual image Q ₂ formed by the half mirror surface 3a with respect to Q ₁ are (-x ₂ , y ₂ , z ₂ ). Therefore, the virtual image of Q ₁
To express Q ₂ in (x, y, z) coordinates, Q ₁
Coordinate transformation of (x, y, z) to (x ₂ , y ₂ , z ₂ ),
What is necessary is to set the x ₂ coordinate to -x ₂ and further transform it back to the (x, y, z) coordinate. Point Q ₂ (x, y, z)
If the value expressed in coordinates is (,,), it is expressed by the following equation.

If we rearrange this formula, we get the following.

As shown in Figure 6, if the angle between the straight line connecting the virtual images Q ₂ and O ₁ with the z-axis when viewed from the x direction is γ, and the angle with the z-axis when viewed from the y direction is δ, then γ,
δ is calculated using the following formula.

From this equation, a ray parallel to the ray 41 passing through Q ₁ O ₁ is a ray parallel to the ray passing through Q ₂ O ₁ , that is,
It can be seen that the light beam is bent toward the viewfinder side by the light beam in the direction determined by δ and γ. Among these parallel rays, if an imaging point is determined by using the ray 42 incident toward the first principal point K1 of the viewfinder master lens ₄ as a representative, the other light beams will also converge to that point. The light ray 42 that has entered the first principal point K ₁ is emitted from the second principal point K ₂ of the viewfinder master lens 4 as a straight line parallel to the incident light ray. The emitted light beam 42 is reflected by the mirror 5 and reaches the first image forming surface 6 of the viewfinder. The intersection of the imaging plane 6 and the light ray 42 becomes the imaging point Q of the parallel rays including the rays 41 and 42.

As shown in Fig. 7, VF master optical axis B ₁ −B ₂
The origin is the intersection O ₂ of the VF optical axis C ₁ −C ₂ , and the VF
Coordinate axis x ^* on optical axis C ₁ − C ₂ , VF master optical axis B ₁ −
Consider coordinates (x ^* , y * , z ^* ) on B ₂ with coordinate axis z ^* and axis y ^* perpendicular to x ^{* and} z ^* , respectively. The direction of the light beam reflected by the mirror 5 is the direction of the straight line connecting the virtual image K ₂ ' of the second principal point K ₂ of the viewfinder master lens 4 by the mirror 5 and the intersection Q ₄ with the mirror surface. Also, the angle between this straight line and the x ^* axis when viewed from the y ^* axis direction, and the angle between the x* axis and the x ^* axis when viewed from the z ^* axis direction are δ and γ, respectively, and the image forming point on the image plane is The position (y ^* , z ^* ) is expressed by the following equation.

y ^* = fm tanγ...(5) z ^* = fm tanδ...(6) By substituting equations (3) and (4) into these equations, the following equations are obtained.

Point Q ₁ is on the zoom optical axis A ₁ −A ₂ , and Q ₁ O ₁
When the prism 3 is not tilted, the rays passing through the lens and their parallel rays, that is, the rays in the direction of the zoom optical axis A ₁ -A ₂ , are connected to the viewfinder first imaging surface 6 when the prism 3 is not tilted.
Since the image is formed at the intersection with the VF optical axis _C1 - _C2 , the prism 3 rotates around the x-axis and the y _-2 axis, Ψ,
The distance between the position of the imaging point of this parallel ray and the VF optical axis C ₁ -C ₂ when the parallel rays are tilted by (θ-45°) is the optical axis deviation. Point Q ₁ is a point on the zoom optical axis A ₁ −A ₂ as described above, but now its coordinates are (1,
0, 0), the optical axis deviations Δy ^* and Δz ^* are equal to y ^* and z ^* from equations (5-2) and (6-2), and are expressed by the following equation.

As is clear from this equation, Δy ^* is a function only of Ψ and does not depend on θ. Also, when Ψ is small and θ is close to 45°, cosΨ1, and |tan2θ| becomes an extremely large value, so cosΨ≪|tan2θ|, and therefore equation (6-3) becomes as follows. Become.

From equation (7), prism 3 is placed on the x axis (zoom optical axis A ₁
-A ₂ ), it can be seen that when the rotation is small, the optical axis moves mainly horizontally. It can also be seen that if the prism 3 is rotated around the _y2 axis, the optical axis will mainly move vertically if the rotation is small. Since the optical axis of the prism 3 does not change even if the prism 3 is moved in parallel, the same can be said even if the above rotation axis is moved in parallel.

Next, consider a plane that includes the x and y axes, and place two points on this plane.
Consider Q ₁ and P ₁ . P ₁ (not shown) is Q ₁ O ₁
Assume that it is not on a straight line passing through. When prism 3 is in the normal position, the ray passing through Q ₁ O ₁ and
The light beam passing through P ₁ O ₁ is imaged on the viewfinder first imaging plane. If these imaging points are Q and P, respectively, the straight line QP becomes a straight line parallel to the y ^* axis. However, if the prism 3 is tilted, the straight line QP will be tilted with respect to the y ^* axis, as shown in FIG. This is called screen tilt. Let the coordinates of Q be (y ^* ₁ , z ^* ₁ ) and the coordinates of P be (y ^* ₂ ,
z ^* ₂ ), the screen tilt H is given by the following formula.

H=tan ^-1 z ^* / ₂ -z ^* / ₁ /y ^* / ₂ -y ^* / ₁ ...(8) Now give (1, 0, 0) to the coordinates of Q and (1, 0, 0) as the coordinates of P. -1, 0), Q is obtained from equation (6-3) and P is given by the following equation.

When Ψ is small and θ is close to 45°, this approximate solution can be expressed by the following equation.

y ^* ₁ = −0.01745fmΨ z ^* ₁ = −0.0349fm (45−θ) H = −Ψ …(10) Next, prism 3 is fabricated and assembled according to the design value, but mirror 5 is tilted. Let's think about the case where there is. After the light coming from the subject passes through the ahocal system of the zoom lens 2, part of the light is separated by the prism 3 to the vidicon 23 for photoelectric conversion, and the other part to the viewfinder side. The image is focused on a first imaging point 6. For incident light parallel to the zoom optical axis _A1 - _A2 , the light is imaged at a point _Q0 on the VF optical axis _C1 - _C2 (see Figure 2).

However, when the mirror 5 is tilted, as shown in FIG. 9, the image is formed at a point Q separated by Δ from the VF optical axis C ₁ -C ₂ . As shown in Figure 10,
Let Q ₅ be a point on the light ray 43 that coincides with the VF master optical axis B ₁ -B ₂ . The virtual image of Q ₅ by the mirror 5 is represented by Q ₅ '. Ray 43 passes through Q ₅ and becomes ray 43 and
It is reflected toward the eyepiece 11a by the surface of the mirror 5 at the intersection _O2 with the VF optical axis _C1 - _C2 . The direction of this reflection is the direction of the line connecting Q ₅ ′O ₂ . Therefore, the inclination δ of this light ray 43 with respect to the VF optical axis C ₁ −C ₂ is
Inclination (θ−45°) with respect to the reference angle of 45° and point Q ₅
It can be seen that it can be expressed as a function of . Here, θ is the angle formed between the VF master optical axis B ₁ -B ₂ and the mirror 5. The intersection of this light ray 43 and the viewfinder first imaging plane becomes the imaging point of the light ray 43 passing through Q ₅ O ₂ . From this, the optical axis deviation Δ can be determined.
The above discussion has been made in two dimensions, but the same applies to three dimensions.

Below, we will derive a formula for calculating the optical axis deviation based on the above-mentioned concept.

As shown in Figure 11, with point O ₂ as the origin, VF
x ^* axis on optical axis C ₁ −C ₂ , VF master optical axis B ₁ −
Rotate the x ^* , y ^* , and z ^* coordinates by an angle Ψ around the x ^* axis, with the z* axis on B ₂ and the y ^* axis perpendicular to the x ^* and ^{z *} axes, respectively, and then The coordinates obtained by rotating the y ^* ₂ axis by an angle θ after rotation are x ^* ₂ ,
Let y ^* ₂ , z ^* ₂ coordinates. Also, the surface of mirror 5 is y ^* ₂ axis
Let z ^* be the plane containing the ₂ axes. Now point Q ₅ in space is x ^* ,
It is expressed as (x ^{* ,} y ^* , z ^* ) using y * , z ^* coordinates, and then the coordinates are transformed and expressed as (x ^* ₂ , y ^* ₂ , z ^* ₂ ). At this time, the virtual image Q ₅ ' of the point Q ₅ with respect to the mirror 5 is (−x ^* ₂ , y ^* ₂ ,
z ^* ₂ ). Transforming the coordinates of this Q ₅ ′ again, we obtain x ^* ,
What is expressed as y ^* , z ^* coordinates ( ^* , ^* ,
^* ), this value is expressed by the following formula.

If we rearrange this equation, we get the following equation.

As shown in Figure 10, a ray passing through Q ₅ O ₂ is
It is reflected by mirror 5, bent in the direction of Q ₅ 'O ₂ and reaches image plane 6. The intersection of this light ray 43 and the viewfinder first imaging surface 6 is the intersection of this light ray 43
This is the focal point of the light beam containing the . z ^* from Q ₅ to O ₂
Let the distance in the axial direction be 1, and let the distance in the z ^* axis direction from O ₂ to Q ₆ be l. However, Q ₆ here is the image forming point when the mirror 5 is not present. In this case, if the position of the imaging point Q is expressed as y ^* , z ^* coordinates and Q=(Y ^* , Z ^* ), this can be expressed by the following equation.

Point Q ₅ is a point on the VF master optical axis B ₁ −B ₂ ,
Ray passing through Q ₅ O ₂ i.e. VF master optical axis B ₁ −
B The rays in _two directions are, if mirror 5 is not tilted,
VF optical axis C ₁ − of viewfinder first imaging point 6
Image is formed at a point on C ₂ . However, when the mirror 5 is tilted by Ψ and (θ-45°) around the x ^* axis and the y ^* ₂ axis, respectively, the VF optical axis C ₁ − of the imaging point of this ray
The distance from C ₂ is the optical axis deviation. If (0, 0, -1) is given as the coordinates of the current point Q ₅ , equations (12) and (13)
The optical axis deviation (Y ^* , Z ^* ) is given by the following equation.

Y ^* Z ^* = −l−cosΨsinΨ(1−cos2θ ⁾ −sin ² Ψ−cos ² Ψcos2θ ^… (14) Next, two points Q ₅ , P ₅ (P ₅ is (not shown), and let the image points of the rays passing through Q ₅ O ₂ and P ₅ O ₂ be Q and P, respectively. When the mirror 5 is not tilted, the straight line QP is parallel to the y ^* axis. It becomes a straight line. However, if the mirror 5 is tilted, the straight line QP will no longer be parallel to the y ^* axis.
The inclination with respect to the y ^* axis at this time becomes the screen inclination (see Fig. 8). Let the coordinates of Q and P be (Y ^* ₁ ,
Z ^* ₁ ), (Y ^* ₂ , Z ^* ₂ ), and if Q ₅ and P ₅ are (0, 0, -1) and (0, -1, -1), respectively, then equation (14) is obtained. From this, Y ^* ₁ and Z ^* ₁ can be found, and Y ^* ₂ Z ^* ₂ can be found using the following formula.

Y ^* Y ^* Z ^* = −l−cos ² Ψ−sin ² Ψcos2θ−cosΨsinΨ(1−co
s2θ) Y ^* Z ^* = −l−cos ² Ψ−sin ² Ψcos2θ−cosΨsinΨ(1−co
s2θ) −sinΨcosΨ(1−cos2θ)−sin ² Ψ−cos ² Ψcos2θ…
(15) The screen inclination H is expressed by the following formula.

H=tan ^-1 Z ^* / ₂ -Z ^* / ₁ /Y ^* / ₂ -Y ^* / ₁ = tan ^-1 sinΨcosΨ (1-cos2θ) / cos ² Ψ＋sin ²
Ψcos2θ (16) When Ψ is small and θ is close to 45°, the following approximate solutions can be obtained from the above equations.

Y ^* ₁ = 0.01745l (1-cos2θ) Ψ Z ^* ₁ = 0.0349l cos ² Ψ (45-θ) ... (17) H = (1-cos2θ) Ψ ... (17) Further finding an approximate solution yields the following It is expressed by the formula.

Y ^* ₁ = 0.01745l・Ψ Z ^* ₁ = 0.0349l (45−θ) H = Ψ … (18) As can be seen from equation (18), Y ^* ₁ and H are the influence of Ψ only, and Z ^* It can be seen that ₁ is strongly influenced only by θ.

If the prism 3 and the mirror 5 are tilted at the same time, the optical axis shift and the screen tilt will appear as a plus. Conversely, by controlling the inclinations of both, desired optical axis deviation and screen inclination can be obtained. For example, the required horizontal and vertical deviation of the optical axis is Δy, Δz screen tilt is ΔH,
The optical axis deviation, screen tilt, and rotation angle due to the prism and mirror are respectively (Δy ₁ , Δz ₁ , ΔH ₁ , Ψ ₁ ,
θ ₁ ), (Δy ₂ , Δz ₂ , ΔH ₂ , Ψ ₂ , θ ₂ ), the following equations hold true.

Δy＝Δy ₁ +Δy ₂ …(19) Δz＝Δz ₁ +Δz ₂ …(20) ΔH＝ΔH ₁ +ΔH ₂ …(21) When Ψ is close to 0 and θ is close to 45°, Δz is approximately the same as θ. Δy and ΔH are mainly functions of Ψ. That is, by substituting equations (10) and (18) to express equations (19), (20), and (21), we get the following.

Δy=−0.01745(fmΨ ₁ −lΨ ₂ ) …(22) Δz=−0.0349 {fm(45-θ ₁ )−l(45-θ ₂ )} …(23) ΔH=−Ψ ₁ +Ψ ₂ …(24 ) Now, if l = fm, then Δy = -0.01745 (Ψ ₁ - Ψ ₂ ) fm Δz = 0 ... (25) ΔH = -Ψ ₁ + Ψ ₂ Therefore, if Ψ ₁ = Ψ ₂ , Δy = 0, △H=0
Therefore, the screen tilt and the optical axis shift can be adjusted all at once.

The optical analysis of the behavior of light when the prism and mirror have a certain error with respect to the design value and the analysis of the adjustment method for the viewfinder have been completed. An example of a reflex type finder will be explained. That is, in this example, the reflective surface of the mirror is installed so as to pass through the second principal point of the viewfinder master lens. Therefore, in this example, fm = l, and the above equation (25) holds true, and even if the optical axis of the single-lens reflex viewfinder is misaligned or the screen is tilted, the mirror 5 is set to VF.
By rotating the prism 3 around the optical axis C ₁ -C ₂ or rotating the prism 3 around the zoom optical axis A ₁ -A ₂ , it is possible to simultaneously adjust the screen tilt and the optical axis shift. .

The above embodiments mainly describe a viewfinder that reflects light after passing through a zoom lens (main lens of a camera) with a prism having a half-mirror treated surface and guides it to a viewfinder master lens. The present invention can also be effectively implemented with a point mirror type finder that uses small pieces to reflect part of the light. Furthermore, the present invention can also be effectively implemented when a prism subjected to half mirror processing is used for the light after passing through the viewfinder master lens, instead of using a mirror. Further, although the above-described examples mainly relate to a VTR camera, it goes without saying that the present invention can be effectively implemented with a normal photographic camera, an 8 mm camera, or the like.

The present invention makes it possible to extremely easily adjust the viewfinder by optically analyzing the behavior of light in a single-lens reflex viewfinder and changing the arrangement of mirrors based on the results, thereby reducing the number of man-hours required for camera adjustment. This has great effects, such as greatly reducing the cost of the camera and reducing the cost of the camera.

[Brief explanation of drawings]

Figure 1 shows the results using a single-lens reflex viewfinder.
A diagram showing the general structure of a VTR camera, Figure 2 shows how the optical axis shifts due to the inclination of the prism, Figure 3 shows the conventional viewfinder adjustment method, and Figure 4 shows the mirror. A two-dimensional analysis of the optical axis shift when the prism is normal and only the prism is tilted, Figures 5 to 7 are diagrams showing the three-dimensional analysis method, and Figure 8 explains the screen tilt. Figure, No. 9
Figure 10 is a two-dimensional diagram explaining the optical axis deviation when the prism is regular and the mirror is tilted, Figure 11.
The figure shows the three-dimensional analysis method. 1...Camera, 2...Main lens (zoom lens),
3... Prism, 3a... Half mirror treated surface, 4...
Viewfinder master lens, 5...Mirror, 6...Viewfinder first imaging surface, 11...
Eyepiece section, 11a...Eyepiece, 21...Master lens, 23...Vidicon, 23a...Imaging surface, 31...
Luminous flux, 41, 42, 43... Ray, A ₁ - _{A 2} ... Zoom optical axis, B ₁ - _{B 2} ... VF master optical axis, C ₁ - C ₂ ...
VF optical axis.

Claims (1)

[Claims] 1. A part of the light focused by the main lens of the camera is reflected by a prism, etc. so that its optical axis is approximately
A single-lens reflex viewfinder that bends the optical axis by 90 degrees, passes through the viewfinder master lens, bends the optical axis approximately 90 degrees again using a mirror to make it parallel to the first optical axis, and provides the eyepiece on the optical axis. In,
A single-lens reflex viewfinder, characterized in that the reflective surface of the mirror is installed so as to pass through a second principal point of the viewfinder master lens.